Task:
Given a weighted tree graph and a set of nodes pairs. For each pair (u,v) from set I need to find(effectively) maximum edge between (u,v).
My approach:
Using Tarjan's algoritm for each pair (u,v) we can find lowest common ancestor LCA(u,v) = a. Then we can present the path between (u,v) as union of (u,a) and (v,a) pathes and maximum edge between (u,v) as max(max_edge(u,a),max_edge(v,a)).
Problem:
I'm trying to add max_edge save in LCA algoritm, but haven't got any success yet.
The question is: How can I add support of max edge save at LCA Tarjan's algorithm?
My attempt code:
int max_cost;
int dsu_find(int node)
{
if (node == parent[node])
return node;
max_cost = std::max(max_cost, edges[node][parent[node]]);
return parent[node] = dsu_find(parent[node]);
}
void lca_dfs(int node, std::vector<std::list<int>> &query_list)
{
dsu_make(node);
ancestor[node] = node;
marks[node] = true;
for(auto neighbour:adjacency_list[node])
{
if (!marks[neighbour.first])
{
lca_dfs(neighbour.first,query_list);
dsu_unite(node, neighbour.first);
ancestor[dsu_find(node)] = node;
}
}
for (auto query_node : query_list[node])
if (marks[query_node])
{
dsu_find(query_node);
dsu_find(node);
printf("%d %d -> %lld\n", node, query_node,max_cost);
query_list[query_node].remove(node);
max_cost = 0;
}
}
But it's working incorrect.
My full lca implementation(without incorrect modifications):
std::vector<int> parent;
std::vector<int> rank;
std::vector<int> ancestor;
std::vector<bool> marks;
std::vector<std::list<std::pair<int, long long>>> adjacency_list;
void lca_dfs(int node, std::vector<std::list<int>> &query_list)
{
dsu_make(node);
ancestor[node] = node;
marks[node] = true;
for(auto neighbour:adjacency_list[node])
{
if (!marks[neighbour.first])
{
lca_dfs(neighbour.first,query_list);
dsu_unite(node, neighbour.first);
ancestor[dsu_find(node)] = node;
}
}
for (auto query_node : query_list[node])
if (marks[query_node])
{
printf("LCA of %d %d is %d\n", node, query_node,ancestor[dsu_find(query_node)]);
query_list[query_node].remove(node);
}
}
//dsu operations
void dsu_make(int node)
{
parent[node] = node;
rank[node] = 0;
}
int dsu_find(int node)
{
return node == parent[node] ? node : parent[node]=dsu_find(parent[node]);
}
void dsu_unite(int node_1,int node_2)
{
int root_1 = dsu_find(node_1), root_2 = dsu_find(node_2);
if(root_1!=root_2)
{
if(rank[root_1] < rank[root_2])
std::swap(root_1, root_2);
parent[root_2] = root_1;
if (rank[root_1] == rank[root_2])
rank[root_1]++;
}
}
*For each node query_list[node] consists of v such as (node,v) is needed pair.
I understood, that I using double memory(just for easier access).
I'll be gratefull for any hints or implementation fixes.
Hope this Implementation works for you.
#include <bits/stdc++.h>
#include <algorithm>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define ass 1e18
#define MOD 1000000007
#define mp make_pair
#define pb push_back
#define pf push_front
#define pob pop_back
#define pof pop_front
#define fi first
#define se second
#define sz(x) (ll)x.size()
#define present(c,x) ((c).find(x) != (c).end())
#define boost ios_base::sync_with_stdio(false);cin.tie(NULL);
#define debug(x) cout << #x << ": " << x << endl;
#define debug2(x,y) cout<<#x<<": "<< x<< ", "<< #y<< ": "<< y<< endl;
#define debug3(x,y,z) cout<<#x<<": "<< x<< ", "<< #y<< ": "<< y<<" "<<#z<<" : "<<z<< endl;
using namespace std;
typedef long long int ll;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update>
pair<int,int> parent[200005],par[200005],dpp[200005][19];
vector<pair< pair<int,int>,pair<int,int> > >v;
vector<pair<int,int> >vv[200005];
int level[200005],vis[200005];
ll ans[200005];
void dfs(int x,int p)
{
level[x]=level[p]+1;
for(int i=0;i<sz(vv[x]);i++)
{
if(vv[x][i].fi!=p)
{
par[vv[x][i].fi].fi=x;
par[vv[x][i].fi].se=vv[x][i].se;
dfs(vv[x][i].fi,x);
}
}
}
void computeparent(int n)
{
for(int i=1;i<=n;i++)
dpp[i][0]=par[i];
for(int j=1;j<=18;j++)
{
for(int i=1;i<=n;i++)
{
dpp[i][j].fi=dpp[dpp[i][j-1].fi][j-1].fi;
dpp[i][j].se=max(dpp[i][j-1].se,dpp[dpp[i][j-1].fi][j-1].se);
}
}
}
int lca(int a,int b)
{
if(level[b]>level[a])
swap(a,b);
int diff=level[a]-level[b];
int m=ceil(log2(diff));
for(int i=m;i>=0;i--)
{
if(diff&(1LL<<i))
a=dpp[a][i].fi;
}
if(a==b)
return a;
for(int i=m;i>=0;i--)
{
if(dpp[a][i].fi!=dpp[b][i].fi)
{
a=dpp[a][i].fi;
b=dpp[b][i].fi;
}
}
return dpp[a][0].fi;
}
int lca2(int a,int b)
{
int c=a,d=b;
if(level[b]>level[a])
swap(a,b);
int i,maxi=0,diff=level[a]-level[b];
int m=ceil(log2(diff));
for(int i=m;i>=0;i--)
{
if(diff&(1LL<<i))
{
maxi=max(maxi,dpp[a][i].se);
a=dpp[a][i].fi;
}
}
return maxi;
}
int finds(int a)
{
while(parent[a].fi!=a)
{
a=parent[a].fi;
}
return a;
}
void unions(int x,int y)
{
if(parent[x].se>parent[y].se)
parent[y].fi=x;
else if(parent[x].se<parent[y].se)
parent[x].fi=y;
else
{
parent[x].fi=y;
parent[x].se++;
}
}
void solve()
{
ll sum=0;
int n,m,i,a,b,c;
cin>>n>>m;
for(i=1;i<=n;i++)
{
parent[i].fi=i;
parent[i].se=0;
}
for(i=0;i<m;i++)
{
cin>>a>>b>>c;
v.pb(mp(mp(c,i),mp(a,b)));
}
sort(v.begin(),v.end());
for(i=0;i<sz(v);i++)
{
int a=v[i].se.fi,b=v[i].se.se;
int x=finds(a),y=finds(b);
if(x!=y)
{
vv[a].pb(mp(b,v[i].fi.fi));
vv[b].pb(mp(a,v[i].fi.fi));
ans[v[i].fi.se]=1;
unions(x,y);
sum+=v[i].fi.fi;
}
}
dfs(1,0);
computeparent(n);
for(i=0;i<m;i++)
{
if(ans[v[i].fi.se]==0)
{
int a=lca(v[i].se.fi,v[i].se.se);
ans[v[i].fi.se]=sum+v[i].fi.fi-max(lca2(a,v[i].se.fi),lca2(a,v[i].se.se));
}
else
ans[v[i].fi.se]=sum;
}
for(i=0;i<m;i++)
cout<<ans[i]<<"\n";
}
int main()
{
boost
int t=1;
//cin>>t;
while(t--)
{
solve();
}
return 0;
}
Related
I have implemented the Weighted graph along with BFS and DFS. But I cannot figure a way out how to stop the traversal when a destination node (specified by user) is reached. Like user should enter the src and dest, and the BFS and DFS algorithm should print the tree until that specified node is reached. I have tried some things but I just cannot understand how to do this. I am attaching the code, any help would be appreciated.
#include "iostream"
#include "vector"
#include "queue"
#include "stack"
using namespace std;
typedef pair<int , int> Pair;
struct Edge{
int src, dest, weight;
};
class Graph{
public:
vector<vector<Pair>> adjacencyList;
Graph(vector<Edge> const &edges, int N)
{
adjacencyList.resize(N);
for(auto &edge: edges)
{
int src = edge.src;
int dest = edge.dest;
int weight = edge.weight;
adjacencyList[src].push_back(make_pair(dest,weight));
adjacencyList[dest].push_back(make_pair(src,weight));
}
}
};
void BFS(Graph const &graph, int src, vector<bool> &discovered)
{
queue<int> q;
discovered[src] = true;
q.push(src);
while(!q.empty())
{
src = q.front();
q.pop();
cout<<src<<" ";
for(int i = 0; i != graph.adjacencyList[src].size() ;i++)
{
if(!discovered[i])
{
discovered[i] = true;
q.push(i);
}
}
}
}
void DFS(Graph const &graph, int src, vector<bool> &discovered)
{
stack<int> stack;
stack.push(src);
while(!stack.empty()){
src = stack.top();
stack.pop();
if(discovered[src])
{
continue;
}
discovered[src] = true;
cout<<src<< " ";
for(int i = 0 ; i < graph.adjacencyList[src].size() ; i++)
{
if(!discovered[i])
{
stack.push(i);
}
}
}
}
void printGraph(Graph const &graph, int N)
{
for (int i = 0; i < N; ++i) {
for(Pair v: graph.adjacencyList[i])
{
cout<<"("<<i<<" , "<<v.first<<" , "<<v.second<<")";
}
cout<<endl;
}
}
int main()
{
vector<Edge> edges =
{
// `(x, y, w)` —> edge from `x` to `y` having weight `w`
{0,1}, {0,2}, {0,3},
{1, 2}, {2,4}, {3,3}, {4,4}
};
int N = 5;
Graph graph(edges,N);
// printGraph(graph,N);
vector<bool> discovered(N, false);
for(int i = 0; i<N; ++i)
{
if(!discovered[i])
{
BFS(graph, i, discovered);
}
}
cout<<endl;
vector<bool> discovered2(N, false);
for(int i = 0; i<N; i++)
{
if(!discovered2[i])
{
DFS(graph, i , discovered2);
}
}
cout<<endl;
printGraph(graph, N);
}
A recursive design makes this much simpler. here is the depth first version
// set stopNode global
......
bool cPathFinder::depthRecurse(int v)
{
// remember this node has been visted
visted[v] = true;
// is this the sop npde
if ( v == stopNode ) {
return true;
}
// look for new adjacent nodes
for (int w : myGraph.all_neighbors(v)) {
if (!visited[w])
{
// search from new node
if( depthRecurse(w) )
return true;
}
}
}
I have to make a program with a hash table that maps single random characters into the table. The program kind of works but sometimes it crashes, also it doesn't map every element. Some of them just won't get inside the table and there are always spare spaces in the table. I don't know what to do to solve these 2 problems. I used 3 versions of open adressing and each of them causes the same 2 problems. Sorry for my bad English. Thank you in advance.
Edited. Of course, I forgot about dynamic allocation. But the problem isn't solved.
#include <time.h>
#include <string>
#include <cstdlib>
using namespace std;
int Liniowo(int i, int proby, int rozmiar) // (open adressing, Linear probing)
{
if(i+proby<rozmiar)
return i+proby;
else
{
return -1;
}
}
int Kwadratowo(int i, int proby, int rozmiar) // (open adressing, Quadratic probing)
{
if (i+proby*proby<rozmiar)
return i+proby*proby;
else
{
return -1;
}
}
int Podwojnie(int i, int proby, int rozmiar, char klucz) // (open adressing, Double hashing)
{
if (i*(klucz*(701%klucz)-klucz%13)<rozmiar&&i*(klucz*(701%klucz)-klucz%13)>0)
return i*(klucz*(701%klucz)-klucz%13);
else
{
return -1;
}
}
int modularnie(char c,int rozmiar) // modular
{
return c%rozmiar;
}
void dodaj(char *tab,int max, char c) // add an element
{
int i=modularnie(c, max);
if (tab[i]== '\0')
tab[i]=c;
else
{
int u=0;
int h;
while (tab[i]!= '\0'&&h!=-1)
{
u++;
// h=Kwadratowo(i, u, max);
h=Podwojnie(i,u,max,c);
}
if (h!=-1)
tab[h]=c;
else
cout << "no niestety, nie udalo sie wstawic " <<endl; //"I couldn't map the element"
}
}
int wyszukaj(char *tab,int max, char c) // search an element
{
int i=modularnie(c, max);
int j=i;
if (tab[i]== '\0')
return -1;
while (tab[i]==c)
{
i=(i+1)%max;
if((i==j)||(tab[i]== '\0'))
return -1;
}
return i;
}
int usun(char *tab,int max, char c) // remove an element
{
int r,j,i=wyszukaj(tab,max,c);
j=i;
if (i==-1)
return -1;
tab[i]= '\0';
while (tab[(++i)%max]!= '\0')
{
i%=max;
r=modularnie(tab[i],max);
if (((i<r)&&(r<=j)) || ((r<=j)&&(j<i)) || ((j<i)&&(i<r)))
{
tab[j]=tab[i];
tab[i]= '\0';
j=i;
continue;
}
}
return 0;
}
int main()
{
srand( time( NULL ) );
int ile;
cout << "podaj wielkosc tablicy: "; //"Type the size of the table"
cin >> ile;
char* tab; // EDITED
tab=new char(ile);
for (int n=0; n<ile; n++)
{
tab[n]= '\0';
}
char e;
for (int i=0; i<ile; i++)
{
e='!'+rand()%127;
dodaj(tab, ile, e);
}
for(int j=0; j<ile; j++)
{
cout << j << ", " << tab[j] << endl;
}
return 0;
}
I can't seem to iterate through all the nodes. It goes through only 1 or two nodes and fails to return true for any case. I want the program to iterator through the node multiple times to get all possible outcome, but still be Eulerian circuit. This is just a small example. To determine if it is Eulerian Circuit, it must pass through the edge only once. Whenever i tried the recursive function DFSUtil, it will stop at the first node or the 2nd node.
Example:
start = A
End = C
A connect to B
A connect to C
B connect to A
B connect to C
C connect to B
Result: 2 path
A -> B -> C -> B -> A -> C
A -> B -> A -> C -> B -> C
#include <iostream>
#include <string>
#include <string.h>
#include <sstream>
#include <stdio.h>
#include <stack>
#include <vector>
using namespace std;
unsigned int V;
class node
{
public:
string Enzyme;
vector<node> connection;
node(string Enzyme)
{
this->Enzyme = Enzyme;
}
bool isEulerCircuit();
bool isConnected();
string DFSUtil(unsigned int v,bool visited[]);
void add_edge(node &n)
{
connection.push_back(n);
cout << Enzyme << " connected to " << n.Enzyme << endl;
}
};
string node::DFSUtil(unsigned int v,bool visited[])
{
visited[v] = true;
vector<node>::iterator it;
string res;
for(it = connection.begin(); it != connection.end(); it++)
{
cout << (*it).Enzyme << endl;
if(!visited[v])
{
res+= (*it).Enzyme;
DFSUtil(v,visited);
}
}
return res;
}
bool node::isEulerCircuit()
{
if (isConnected() == false)
{
return false;
}
return true;
}
bool node::isConnected()
{
bool visited[V];
for(int i = 0; i < V; i++)
{
visited[i] = false;
}
int n=3;
DFSUtil(n,visited);
for (int i = 0; i < V; i++)
{
if (visited[i] == false)
{
return false;
}
}
return true;
}
int main()
{
vector<node> nod;
string A = "A";
string B = "B";
string C = "C";
nod.push_back(A);
nod.push_back(B);
nod.push_back(C);
for(int i = 0; i < nod.size(); i++)
{
V = i;
}
cout << endl;
nod[0].add_edge(nod[1]);
nod[0].add_edge(nod[2]);
nod[1].add_edge(nod[0]);
nod[1].add_edge(nod[2]);
nod[2].add_edge(nod[1]);
if(nod[0].isEulerCircuit())
{
cout << "HI" << endl;
}
else
{
cout << "BYE" << endl;
}
return 0;
}
I just learnt Dijkstra's algorithm and solved a few problems and I am trying to solve this http://codeforces.com/problemset/problem/20/C problem but I am getting Time limit Exceeded in a large test case, I want to know if my code can be further optimized in any way or if there is any other faster implementation of Dijkstra then let me know.
My code
#include<bits/stdc++.h>
using namespace std;
#define pii pair<int,int>
#define vp vector<pii>
int p[100010],d[100010];
void printer(int current)
{
if(p[current]==-2)
{
printf("%d ",current);
return;
}
printer(p[current]);
printf("%d ",current);
}
class Prioritize
{
public:
int operator()(const pii &p1,const pii &p2)
{
return p1.second<p2.second;
}
};
int main()
{
priority_queue<pii, vp, Prioritize> Q;
int nv;
scanf("%d",&nv);
vp g[nv+1];
int ne,u,v,w;
scanf("%d",&ne);
for(int i=0;i<ne;i++)
{
scanf("%d %d %d",&u,&v,&w);
g[u].push_back(pii(v,w));
g[v].push_back(pii(u,w));
}
int source=1;
int size;
for(int i=1;i<=nv;i++)
{
d[i]=INT_MAX;
p[i]=-1;
}
d[source]=0;
p[source]=-2;//marker for source.
Q.push(pii(source,d[source]));
while(!Q.empty())
{
u=Q.top().first;
Q.pop();
size=g[u].size();
for(int i=0;i<size;i++)
{
v=g[u][i].first;
w=g[u][i].second;
if(d[v]>d[u]+w)
{
d[v]=d[u]+w;
p[v]=u;
Q.push(pii(v,d[v]));
}
}
}
/*for(int i=1;i<=nv;i++)
{
printf("Node %d and min weight = %d and parent = %d\n",i,d[i],p[i]);
}*/
if(p[nv]==-1)
{
printf("%d\n",-1);
return 0;
}
printer(nv);
return 0;
}
Your algorithm has complexity O(nm). I propose to add the following
Q.push(pii(source,d[source]));
while(!Q.empty())
{
u=Q.top().first;
int curD = Q.top().second; //this
if( curD > d[u]) continue; //and this
Q.pop();
size=g[u].size();
for(int i=0;i<size;i++)
{
After changes complexity will be O(mlogn). If you now russian you can read http://e-maxx.ru/algo/dijkstra_sparse
why my program Time Limited Error?
because of the sort?
this is the question
link text
#include <cstdio>
#include <cstring>
using namespace std;
int map[22][44];
int book[22];
int total;
int sum;
int way[22];
int tails[22];
int tail;
void init()
{
memset(map,0,sizeof(map));
memset(book,0,sizeof(book));
sum =0;
memset(way,0,sizeof(way));
way[1]=1;
memset(tails,0,sizeof(tails));
}
void sort()
{
int t;
for (int i=1;i<=22;i++)
{
if (tails[i]==0)
break;
else
{
for (int j=1;j<=tails[i]-1;j++)
for (int k=j+1;k<=tails[i];k++)
{
if (map[i][j] > map[i][k])
{
t = map[i][j];
map[i][j]=map[i][k];
map[i][k]=t;
}
}
}
}
}
void dfs(int x,int y)
{
if ((x < 1)||(x > 22))
return;
if (book[x]==1)
return;
//printf("%d \n",x);
if (x == total)
{
sum++;
for (int i=1;i<=y-1;i++)
{
printf("%d ",way[i]);
}
printf("%d",total);
printf("\n");
return;
}
tail = tails[x];
for (int i=1;i<=43;i++)
{
book[x]=1;
way[y]=x;
dfs(map[x][i],y+1);
book[x]=0;
}
}
int main()
{
int temp1,temp2;
//freopen("ex.in","r",stdin);
//freopen("ex.out","w",stdout);
int c = 0;
while(scanf("%d",&total)!=EOF)
{
c++;
printf("CASE ");
printf("%d",c);
printf(":");
printf("\n");
init();
for (;;)
{
scanf("%d%d",&temp1,&temp2);
if ((temp1 == 0)&&(temp2 == 0))
break;
else
{
tails[temp1]++;
tail = tails[temp1];
map[temp1][tail]=temp2;
tails[temp2]++;
tail = tails[temp2];
map[temp2][tail]=temp1;
}
}
sort();
dfs(1,1);
printf("There are ");printf("%d",sum);printf(" routes from the firestation to streetcorner ");printf("%d",total);printf(".");
printf("\n");
}
return 0;
}
Because your sorting algoritm is in worst-case O(n*n), you can use InnoSort for better worst-case complexity O(n*log(n)).
You are using C++ then use sort function from <algorithm> header to do this simplest.
Documentation you can find at http://www.sgi.com/tech/stl/sort.html
For a start, you're accessing tails and map past the end. C++ arrays are zero-indexed, so the first element is 0, and the last valid elements are tails[21] and map[21][43].