In the code below:
#define MAX_VERTICES 260000
#include <fstream>
#include <vector>
#include <queue>
#define endl '\n'
using namespace std;
struct edge {
int dest;
int length;
};
bool operator< (edge e1, edge e2) {
return e1.length > e2.length;
}
int C, P, P0, P1, P2;
vector<edge> edges[MAX_VERTICES];
int best1[MAX_VERTICES];
int best2[MAX_VERTICES];
void dijkstra (int start, int* best) {
for (int i = 0; i < P; i++) best[i] = -1;
best[start] = 0;
priority_queue<edge> pq;
edge first = { start, 0 };
pq.push(first);
while (!pq.empty()) {
edge next = pq.top();
pq.pop();
if (next.length != best[next.dest]) continue;
for (vector<edge>::iterator i = edges[next.dest].begin(); i != edges[next.dest].end(); i++) {
if (best[i->dest] == -1 || next.length + i->length < best[i->dest]) {
best[i->dest] = next.length + i->length;
edge e = { i->dest, next.length+i->length };
pq.push(e);
}
}
}
}
int main () {
ifstream inp("apple.in");
ofstream outp("apple.out");
inp >> C >> P >> P0 >> P1 >> P2;
P0--, P1--, P2--;
for (int i = 0; i < C; i++) {
int a, b;
int l;
inp >> a >> b >> l;
a--, b--;
edge e = { b, l };
edges[a].push_back(e);
e.dest = a;
edges[b].push_back(e);
}
dijkstra (P1, best1); // find shortest distances from P1 to other nodes
dijkstra (P2, best2); // find shortest distances from P2 to other nodes
int ans = best1[P0]+best1[P2]; // path: PB->...->PA1->...->PA2
if (best2[P0]+best2[P1] < ans)
ans = best2[P0]+best2[P1]; // path: PB->...->PA2->...->PA1
outp << ans << endl;
return 0;
}
What is this: if (next.length != best[next.dest]) continue; used for? Is it to avoid us situations where going through the loop will give us the same answer that we already have?
Thanks!
That line is a way to handle the fact that c++'s priority_queue does not have a decrease_key function.
That is, when you do pq.push(e) and there is already an edge with the same destination in the heap you would prefer to decrease the key of the edge already in the heap. This is not easily done with c++'s priority_queue and so a simple way to handle it is to allow multiple edges in the heap corresponding to the same destination and ignoring all but the first (for each dest) that you pop from the heap.
Note that this changes the complexity from O(ElogV) to O(ElogE).
I guess you are contemplating the case where your priority_queue contains 2 times the same edge, but each one with a different "length".
This could happen if you push edge X which has a length of Y, and afterwards push edge X again, but this time it has a length < Y. That is why, if the length of that edge, isn't the lowest you've found for that edge so far, you ommit it in that loop's iteration.
Related
I'm attempting a problem on SPOJ where I'm supposed to find the longest path between any two nodes in a tree. The input consists of the number of testcases t, the number of nodes n, followed by n-1 edges given by "a b l", with a referring to node 1, b referring to node 2, and l referring to the length of the edge. I tried using the double dfs method, performing dfs on node 1 first to find the longest possible path beginning from node 1. Afterwards, I perform dfs on the node furthest away from node 1, finding the longest possible distance. Unfortunately, my code is wrong and I'm not sure why at all, and i'm hoping someone could help me out. Thanks in advance!
EDIT: Forgot to mention that I did manage to solve the problem using double BFS. I wanted to try using DFS to solve it as well because DFS is supposedly easier to implement than BFS, but using DFS gives me the wrong answer.
#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int>> adj[50001];
bool visited[50001] = {0};
int maxdist = -1, maxnode = -1;
void dfs(int node, int d)
{
visited[node] = 1;
if (d > maxdist)
{
maxdist = d;
maxnode = node;
}
for(auto u: adj[node])
{
if(!visited[u.first])
{
dfs(u.first, d+u.second);
}
}
return;
}
int main()
{
int t;
cin >> t;
while(t--)
{
int n;
cin >> n;
for (int i = 0; i < n-1; i++)
{
int a, b, l;
cin >> a >> b >> l;
adj[a].push_back(make_pair(b, l));
adj[b].push_back(make_pair(a, l));
}
dfs(1, 0);
for(int i = 1; i <= n; i++)
{
visited[i] = 0;
}
dfs(maxnode, 0);
cout << maxdist << endl;
}
}
Not sure if this will make it work, but you are sharing adj[] and visited[] between different test cases, make sure to clean them up before a new test case.
So i was trying to make the challage: Breadth First Search: Shortest Reach on HackerRank, but i keep getting the bad alloc exception when the tests have great numbers of node/edges. The program works on the first test, so i don't think, it's something wrong with the implementation.
So here is the implementation:
(sorry for the indentation , my first question)
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <queue>
#include <limits.h>
using namespace std;
int main() {
//test numbers
int t;
//the cost
int cost = 6;
cin >> t;
//for each test
for (int nt = 0; nt < t; ++nt) {
int n, e;
int snode;
queue <int> *que = new queue<int>();
//read the node/edges
cin >> n >> e;
//distance/visited/parents arrays/adjlist vector
int dist[n + 1] = {-1};
bool visited[n + 1] = {false};
int parents[n + 1] = {-1};
vector< vector<int> > adjList(n + 1);
//read into the adjlist, unoriented graph, each edge has 6 weight
for (int ne = 0; ne < e; ++ne) {
int x, y;
cin >> x >> y;
adjList[x].push_back(y);
adjList[y].push_back(x);
}
//read the starting node
cin >> snode;
dist[snode] = 0;
//do actual bfs
que->push(snode);
visited[snode] = true;
while(!que->empty()) {
int c_node = que->front();
que->pop();
for (int i = 0; i < adjList[c_node].size(); ++i) {
if (visited[adjList[c_node].at(i)] == false) {
que->push(adjList[c_node].at(i));
parents[adjList[c_node].at(i)] = c_node;
dist[adjList[c_node].at(i)] = dist[parents[adjList[c_node].at(i)]] + cost;
visited[adjList[c_node].at(i)] == true;
}
}
}
//print at output the distance from the starting node to each other node
//if unreachable, print -1
for (int i = 1; i < n + 1; ++i) {
if (i == snode) {
} else if (dist[i] == 0 && i != snode) {
cout << "-1 ";
} else {
cout << dist[i] << " ";
}
}
cout << "\n";
}
return 0;
}
Am i doing something wrong, i haven't seen anyone else complain on this matter in the discussion section of the site.
How can i avoid the exception to be thrown and from where does it come?
Thank you!
I don't know, exactly, what is the cause of your exception; and I don't know ho to reproduce your problem because depends (I suppose) from the input values. A lot of input values, I suppose.
But I see some weak points (IMHO) of your code, so I try to point your attention to them.
1) you alloc a std::queue in your for cycle
queue <int> *que = new queue<int>();
but you never free it; it's a waste of memory
2) you're using C-style variable-length arrays
int dist[n + 1] = {-1};
bool visited[n + 1] = {false};
int parents[n + 1] = {-1};
They aren't valid C++ standard code. I suggest you the use of standard containers (std::vector or std::queue).
3) you're initializing your C-style variable-length arrays with a initializers lists with only an element (-1 or false). I suppose your intention was initialize all n+1 elements with -1 and false. But this syntax initialize only the first element of the array with -1 and false.
If you want to initialize all n+1 element to -1 and false, the solution is (again) use standard containers; by example
std::vector<int> dist(n+1, -1);
std::vector<bool> visited(n+1, false);
std::vector<int> parents(n+1, -1);
4) you access arrays without bounds checking. By example:
cin >> snode;
dist[snode] = 0;
where snode is a int variable; if you insert a negative value, or a value over n, you write dist out of its bounds, devastating the memory. This, I suppose, can explain your "bad alloc exception".
Suggestion: use standard containers (again) instead of C-style array and use at() (that perform bounds checking) instead []; so
cin >> snode;
dist.at(snode) = 0;
5) sorry for my bad English (ok, I'm joking: this isn't one of your weak points; this is one of mine).
In the input specification N the number of nodes and M the number of edges are given . So the first simple check is that M should be equal to N-1 otherwise it simply can't be a tree.
What I did next was just a DFS in which I see that whether during the DFS we come across a visited a node again ( different from the parent node, by parent node I mean the node which has called the dfs of the next node adjacent to it ) then it means that we have a cycle and it isn't a tree . But apparently my solution keeps on getting a wrong answer . I am posting the code but only the snippets that are important . I am storing the graph as a adjacency list and I am posting the function isTree() which tests whether it is a tree or not ? What is the correct logic ?
#include <iostream>
#include <list>
using namespace std;
// Graph class represents a directed graph using adjacency list representation
class Graph
{
int V; // No. of vertices
list<int> *adj; // Pointer to an array containing adjacency lists
bool isTreeUtil(int v, bool visited[],int parent);
public:
Graph(int V); // Constructor
void addEdge(int v, int w); // function to add an edge to graph
bool isTree(); // Tells whether the given graph is a tree or not
void printGraph();
};
Graph::Graph(int V)
{
this->V = V;
adj = new list<int>[V+1];
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w); // Add w to v’s list.
adj[w].push_back(v);
}
bool Graph::isTreeUtil(int v, bool visited[],int parent)
{
//int s_v = v;
visited[v] = true;
list<int>::iterator i;
for(i = adj[v].begin(); i != adj[v].end(); ++i) {
if (!visited[*i])
isTreeUtil(*i,visited,v);
else {
if (*i != parent && visited[*i])
return false;
}
}
return true;
}
bool Graph::isTree() {
bool *visited = new bool[V+1];
for(int i = 1; i < V+1; i++)
visited[i] = false;
visited[1] = true; // marking the first node as visited
for(int i = 1; i < V+1; i++)
visited[i] = false;
int parent = -1; // initially it has no parent
//list<int> :: iterator i;
//for (i = adj[v].begin(); i != adj[v].end(); ++i)
return isTreeUtil(1, visited, parent);
}
void Graph::printGraph() {
for (int i = 1;i <= this->V; i++) {
cout << i << "->";
list<int>::iterator j;
for (j = adj[i].begin(); j != adj[i].end(); ++j) {
cout << *j << "->";
}
cout << "\n";
}
}
int main() {
int N, M;
cin >> N >> M;
Graph G(N);
int v, w;
int m = 0;
while (m < M) {
cin >> v >> w;
G.addEdge(v,w);
m++;
}
if (M != N-1) {
cout << "NO\n";
else if (G.isTree())
cout << "YES\n";
else
cout << "NO\n";
}
I took your code, compiled, and ran it on my machine. When implementing a graph, there are important specs to consider. When you choose to obey a spec, it is generally good practice to enforce that spec in your code.
It is already clear that the graph has 2-way edges, though it does not hurt to specifically mention this.
Allow Duplicate Edges?
Your program allows me to make edge (1,2) and then another edge (1,2) and count it as 2 edges. This makes your conditional M != N-1 an insufficient check. Either disallow duplicate edges or account for them in your algorithm (currently, a duplicate edge will cause your algorithm to return incorrectly).
Self Edges?
Does your graph allow a vertex to have an edge to itself? If so, should the self-path invalidate the tree (perhaps a self-loop is legal because in a tree, every node can access itself)? Currently, self edges also break your algorithm.
To help you, here is my revised implementation of addEdge() that disallows duplicate edges and disallows self-loops. As a bonus it also checks for array bounds ;)
Please note that the additional include, and the change in function signature (it now returns a bool).
#include <algorithm>
bool Graph::addEdge(int v, int w)
{
// sanity check to keep us from seg faulting
if (v < 1 || v > this->V || w < 1 || w > this->V) {
return false;
}
// no self-edges
if (w == v) {
return false;
}
// no duplicate edges allowed either
std::list<int>::iterator findV = std::find(adj[v].begin(), adj[v].end(), w);
std::list<int>::iterator findW = std::find(adj[w].begin(), adj[w].end(), v);
if (findV != adj[v].end() || findW != adj[w].end()) {
return false;
}
adj[v].push_back(w); // Add w to v’s list.
adj[w].push_back(v);
return true;
}
I hope this helps. If this is an assignment, you should review the write-up. They must have specified these cases if your implementation was auto-graded. As #congusbongus mentioned, your algorithm also fails in the case of a disconnected node.
Please note that you also have to revise the main() method in order for my implementation to work. Change this part of the function:
while (m < M) {
cout << "Create Edge from x to y" << endl;
cin >> v >> w;
if (!G.addEdge(v,w)) {
cout << ">>Invalid edge not added" << endl;
} else {
cout << ">>Successfully added edge" << endl;
m++;
}
}
it runs for all the simple test cases I drew on paper but when submitting it fails !
Sounds like some auto-marking system for homework right? If you had access to the exact test cases, then the problem would be obvious. In this case it's probably not available, so we can only speculate.
In my experience, most failures of this kind are due to missed boundary cases. You say you check for number of edges = number of nodes - 1, but have you also considered the following?
All nodes connected
No more than one edge per pair of nodes
That is, is your program prepared to return "NO" for this?
_
/ \
o o---o
Nodes: 3, edges: 2
We have a problem here, we're trying to find all the shortest paths in graph from one node to another. We have already implemented dijkstra but we really dont know how to find them all.
Do we have to use BFS?
#include <vector>
#include <iostream>
#include <queue>
using namespace std;
typedef pair <int, int> dist_node;
typedef pair <int, int> edge;
const int MAXN = 10000;
const int INF = 1 << 30;
vector <edge> g[MAXN];
int d[MAXN];
int p[MAXN];
int dijkstra(int s, int n,int t){
for (int i = 0; i <= n; ++i){
d[i] = INF; p[i] = -1;
}
priority_queue < dist_node, vector <dist_node>,greater<dist_node> > q;
d[s] = 0;
q.push(dist_node(0, s));
while (!q.empty()){
int dist = q.top().first;
int cur = q.top().second;
q.pop();
if (dist > d[cur]) continue;
for (int i = 0; i < g[cur].size(); ++i){
int next = g[cur][i].first;
int w_extra = g[cur][i].second;
if (d[cur] + w_extra < d[next]){
d[next] = d[cur] + w_extra;
p[next] = cur;
q.push(dist_node(d[next], next));
}
}
}
return d[t];
}
vector <int> findpath (int t){
vector <int> path;
int cur=t;
while(cur != -1){
path.push_back(cur);
cur = p[cur];
}
reverse(path.begin(), path.end());
return path;
}
This is our code, we believe we have to modify it but we really don't know where.
Currently, you are only saving/retrieving one of the shortest paths that you happen to find. Consider this example:
4 nodes
0 -> 1
0 -> 2
1 -> 3
2 -> 3
It becomes clear that you cannot have a single p[] value for each position, as in fact the 4th node (3) has 2 previous valid nodes: 1 and 2.
You could thus replace it with a vector<int> p[MAXN]; and work as follows:
if (d[cur] + w_extra < d[next]){
d[next] = d[cur] + w_extra;
p[next].clear();
p[next].push_back(cur);
q.push(dist_node(d[next], next));
}
else if(d[cur] + w_extra == d[next]){
p[next].push_back(cur); // a new shortest way of hitting this same node
}
You will also need to update your findpath() function as it will need to deal with "branches" resulting in several multiple paths, possibly an exponentially huge amount of paths depending on the graph. If you just need to print the paths, you could do something like this:
int answer[MAXN];
void findpath (int t, int depth){
if(t == -1){ // we reached the initial node of one shortest path
for(int i = depth-1; i >= 0; --i){
printf("%d ", answer[i]);
}
printf("%d\n", last_node); // the target end node of the search
return;
}
for(int i = p[t].size()-1; i >= 0; --i){
answer[depth] = p[t][i];
findpath(p[t][i], depth+1);
}
}
Note you'll need to do p[s].push_back(-1) at the beginning of your dijkstra, besides clearing this vector array between cases.
For example suppose there are 3 nodes A,B,C and A links to B and C, B links to A and C, and C links to B and A. In visual form its like this
C <- A -> B //A links to B & C
A <- B -> C //B links to A & C
B <- C -> A //C links to B & A
Assume the A,B,C are held in an array like so [A,B,C] with index starting at 0. How can I efficiently sort the array [A,B,C] according to the value held by each node.
For example if A holds 4, B holds -2 and C holds -1, then sortGraph([A,B,C]) should return [B,C,A]. Hope its clear. Would it be possible if I can somehow utilize std::sort?
EDIT: Not basic sort algorithm. Let me clarify a bit more. Assume I have a list of Nodes [n0,n1...nm]. Each ni has a left and right neighbor index. For example, n1 left neight is n0 and its right neighbor is n2. I use index to represent the neighbor. If n1 is at index 1, then its left neighbor is at index 0 and its right neighbor is at index 2. If I sort the array, then I need to update the neighbor index as well. I don't want to really implement my own sorting algorithm, any advice on how to proceed?
If I understand the edited question correctly your graph is a circular linked list: each node points to the previous and next nodes, and the "last" node points to the "first" node as its next node.
There's nothing particularly special you need to do the sort that you want. Here are the basic steps I'd use.
Put all the nodes into an array.
Sort the array using any sorting algorithm (e.g. qsort).
Loop through the result and reset the prev/next pointers for each node, taking into account the special cases for the first and last node.
Here is a C++ implementation, hope is useful (it includes several algorithms like dijkstra, kruskal, for sorting it uses depth first search, etc...)
Graph.h
#ifndef __GRAPH_H
#define __GRAPH_H
#include <vector>
#include <stack>
#include <set>
typedef struct __edge_t
{
int v0, v1, w;
__edge_t():v0(-1),v1(-1),w(-1){}
__edge_t(int from, int to, int weight):v0(from),v1(to),w(weight){}
} edge_t;
class Graph
{
public:
Graph(void); // construct a graph with no vertex (and thus no edge)
Graph(int n); // construct a graph with n-vertex, but no edge
Graph(const Graph &graph); // deep copy of a graph, avoid if not necessary
public:
// #destructor
virtual ~Graph(void);
public:
inline int getVertexCount(void) const { return this->numV; }
inline int getEdgeCount(void) const { return this->numE; }
public:
// add an edge
// #param: from [in] - starting point of the edge
// #param: to [in] - finishing point of the edge
// #param: weight[in] - edge weight, only allow positive values
void addEdge(int from, int to, int weight=1);
// get all edges
// #param: edgeList[out] - an array with sufficient size to store the edges
void getAllEdges(edge_t edgeList[]);
public:
// topological sort
// #param: vertexList[out] - vertex order
void sort(int vertexList[]);
// dijkstra's shortest path algorithm
// #param: v[in] - starting vertex
// #param: path[out] - an array of <distance, prev> pair for each vertex
void dijkstra(int v, std::pair<int, int> path[]);
// kruskal's minimum spanning tree algorithm
// #param: graph[out] - the minimum spanning tree result
void kruskal(Graph &graph);
// floyd-warshall shortest distance algorithm
// #param: path[out] - a matrix of <distance, next> pair in C-style
void floydWarshall(std::pair<int, int> path[]);
private:
// resursive depth first search
void sort(int v, std::pair<int, int> timestamp[], std::stack<int> &order);
// find which set the vertex is in, used in kruskal
std::set<int>* findSet(int v, std::set<int> vertexSet[], int n);
// union two sets, used in kruskal
void setUnion(std::set<int>* s0, std::set<int>* s1);
// initialize this graph
void init(int n);
// initialize this graph by copying another
void init(const Graph &graph);
private:
int numV, numE; // number of vertices and edges
std::vector< std::pair<int, int> >* adjList; // adjacency list
};
#endif
Graph.cpp
#include "Graph.h"
#include <algorithm>
#include <map>
Graph::Graph()
:numV(0), numE(0), adjList(0)
{
}
Graph::Graph(int n)
:numV(0), numE(0), adjList(0)
{
this->init(n);
}
Graph::Graph(const Graph &graph)
:numV(0), numE(0), adjList(0)
{
this->init(graph);
}
Graph::~Graph()
{
delete[] this->adjList;
}
void Graph::init(int n)
{
if(this->adjList){
delete[] this->adjList;
}
this->numV = n;
this->numE = 0;
this->adjList = new std::vector< std::pair<int, int> >[n];
}
void Graph::init(const Graph &graph)
{
this->init(graph.numV);
for(int i = 0; i < numV; i++){
this->adjList[i] = graph.adjList[i];
}
}
void Graph::addEdge(int from, int to, int weight)
{
if(weight > 0){
this->adjList[from].push_back( std::make_pair(to, weight) );
this->numE++;
}
}
void Graph::getAllEdges(edge_t edgeList[])
{
int k = 0;
for(int i = 0; i < numV; i++){
for(int j = 0; j < this->adjList[i].size(); j++){
// add this edge to edgeList
edgeList[k++] = edge_t(i, this->adjList[i][j].first, this->adjList[i][j].second);
}
}
}
void Graph::sort(int vertexList[])
{
std::pair<int, int>* timestamp = new std::pair<int, int>[this->numV];
std::stack<int> order;
for(int i = 0; i < this->numV; i++){
timestamp[i].first = -1;
timestamp[i].second = -1;
}
for(int v = 0; v < this->numV; v++){
if(timestamp[v].first < 0){
this->sort(v, timestamp, order);
}
}
int i = 0;
while(!order.empty()){
vertexList[i++] = order.top();
order.pop();
}
delete[] timestamp;
return;
}
void Graph::sort(int v, std::pair<int, int> timestamp[], std::stack<int> &order)
{
// discover vertex v
timestamp[v].first = 1;
for(int i = 0; i < this->adjList[v].size(); i++){
int next = this->adjList[v][i].first;
if(timestamp[next].first < 0){
this->sort(next, timestamp, order);
}
}
// finish vertex v
timestamp[v].second = 1;
order.push(v);
return;
}
void Graph::dijkstra(int v, std::pair<int, int> path[])
{
int* q = new int[numV];
int numQ = numV;
for(int i = 0; i < this->numV; i++){
path[i].first = -1; // infinity distance
path[i].second = -1; // no path exists
q[i] = i;
}
// instant reachable to itself
path[v].first = 0;
path[v].second = -1;
while(numQ > 0){
int u = -1; // such node not exists
for(int i = 0; i < numV; i++){
if(q[i] >= 0
&& path[i].first >= 0
&& (u < 0 || path[i].first < path[u].first)){ //
u = i;
}
}
if(u == -1){
// all remaining nodes are unreachible
break;
}
// remove u from Q
q[u] = -1;
numQ--;
for(int i = 0; i < this->adjList[u].size(); i++){
std::pair<int, int>& edge = this->adjList[u][i];
int alt = path[u].first + edge.second;
if(path[edge.first].first < 0 || alt < path[ edge.first ].first){
path[ edge.first ].first = alt;
path[ edge.first ].second = u;
}
}
}
delete[] q;
return;
}
// compare two edges by their weight
bool edgeCmp(edge_t e0, edge_t e1)
{
return e0.w < e1.w;
}
std::set<int>* Graph::findSet(int v, std::set<int> vertexSet[], int n)
{
for(int i = 0; i < n; i++){
if(vertexSet[i].find(v) != vertexSet[i].end()){
return vertexSet+i;
}
}
return 0;
}
void Graph::setUnion(std::set<int>* s0, std::set<int>* s1)
{
if(s1->size() > s0->size()){
std::set<int>* temp = s0;
s0 = s1;
s1 = temp;
}
for(std::set<int>::iterator i = s1->begin(); i != s1->end(); i++){
s0->insert(*i);
}
s1->clear();
return;
}
void Graph::kruskal(Graph &graph)
{
std::vector<edge_t> edgeList;
edgeList.reserve(numE);
for(int i = 0; i < numV; i++){
for(int j = 0; j < this->adjList[i].size(); j++){
// add this edge to edgeList
edgeList.push_back( edge_t(i, this->adjList[i][j].first, this->adjList[i][j].second) );
}
}
// sort the list in ascending order
std::sort(edgeList.begin(), edgeList.end(), edgeCmp);
graph.init(numV);
// create disjoint set of the spanning tree constructed so far
std::set<int>* disjoint = new std::set<int>[this->numV];
for(int i = 0; i < numV; i++){
disjoint[i].insert(i);
}
for(int e = 0; e < edgeList.size(); e++){
// consider edgeList[e]
std::set<int>* s0 = this->findSet(edgeList[e].v0, disjoint, numV);
std::set<int>* s1 = this->findSet(edgeList[e].v1, disjoint, numV);
if(s0 == s1){
// adding this edge will make a cycle
continue;
}
// add this edge to MST
graph.addEdge(edgeList[e].v0, edgeList[e].v1, edgeList[e].w);
// union s0 & s1
this->setUnion(s0, s1);
}
delete[] disjoint;
return;
}
#define IDX(i,j) ((i)*numV+(j))
void Graph::floydWarshall(std::pair<int, int> path[])
{
// initialize
for(int i = 0; i < numV; i++){
for(int j = 0; j < numV; j++){
path[IDX(i,j)].first = -1;
path[IDX(i,j)].second = -1;
}
}
for(int i = 0; i < numV; i++){
for(int j = 0; j < this->adjList[i].size(); j++){
path[IDX(i,this->adjList[i][j].first)].first
= this->adjList[i][j].second;
path[IDX(i,this->adjList[i][j].first)].second
= this->adjList[i][j].first;
}
}
// dynamic programming
for(int k = 0; k < numV; k++){
for(int i = 0; i < numV; i++){
for(int j = 0; j < numV; j++){
if(path[IDX(i,k)].first == -1
|| path[IDX(k,j)].first == -1){
// no path exist from i-to-k or from k-to-j
continue;
}
if(path[IDX(i,j)].first == -1
|| path[IDX(i,j)].first > path[IDX(i,k)].first + path[IDX(k,j)].first){
// there is a shorter path from i-to-k, and from k-to-j
path[IDX(i,j)].first = path[IDX(i,k)].first + path[IDX(k,j)].first;
path[IDX(i,j)].second = k;
}
}
}
}
return;
}
If you are looking for sorting algorithms you should just ask google:
http://en.wikipedia.org/wiki/Sorting_algorithm
My personal favourite is the BogoSort coupled with parallel universe theory. The theory is that if you hook a machine up to the program that can destroy the universe, then if the list isn't sorted after one iteration it will destroy the universe. That way all the parallel universes except the one with the list sorted will be destroyed and you have a sorting algorithm with complexity O(1).
The best ....
Create a struct like this:
template<typename Container, typename Comparison = std::less<typename Container::value_type>>
struct SortHelper
{
Container const* container;
size_t org_index;
SortHelper( Container const* c, size_t index ):container(c), org_index(index) {}
bool operator<( SortHelper other ) const
{
return Comparison()( (*c)[org_index], (*other.c)[other.org_index] );
}
};
This lets you resort things however you want.
Now, make a std::vector<SortHelper<blah>>, sort it, and you now have a vector of instructions of where everything ends up going after you sort it.
Apply these instructions (there are a few ways). An easy way would be to reuse container pointer as a bool. Walk the sorted vector of helpers. Move the first entry to where it should go, moving what you found where it should go to where it should go, and repeat until you loop or the entire array is sorted. As you go, clear the container pointers in your helper struct, and check them to make sure you don't move an entry that has already been moved (this lets you detect loops, for example).
Once a loop has occurred, proceed down the vector looking for the next as-yet-not-in-right-place entry (with a non-null container pointer).