connected components in undirected graphs in c++ - c++

Question

I wanted to count the number of groups in an undirected graphs in c++.I was trying using bfs but was unsuccessful.I have been given a range of numbers [L,R] (or think of these ranges as number of vertices) and i have to find the number of groups.How do i do this?
Like if i have (Input):
1 3
2 5
6 9
Output:
2
As there are 2 groups.
My code:
bool visited[MAX];
vector<int> v[MAX];
int solve(int x)
{
queue<int> q;int ans=0;
q.push(x);
if(v[x].empty())
{
ans++;
}
while(!q.empty())
{
int curr = q.front();
visited[curr] = true;
q.pop();
for(int i = 0; i < v[curr].size(); i ++)
{
if(!visited[v[curr][i]])
{
q.push(v[curr][i]);
visited[v[curr][i]] = true;
}
}
if(v[curr].empty()) ans++;
}
return ans;
}
int main()
{
int t;scanf("%d",&t);
while(t--)
{
int l,r,n,ans=0,min_,max_=0;
scanf("%d",&n);
for(int i = 0; i < n; i ++)
visited[i] = false;
for(int j=0;j<n;j++)
{
scanf("%d",&l);scanf("%d",&r);
for(int i=l;i<r;i++)
{
v[i].push_back(i+1);
min_ = min(min_,i);
max_ = max(max_,i+1);
}
}
printf("%d\n",solve(min_));
}
return 0;
}

Answer 1

Let's take a look at how many edges are created in the worst case. It is N * (MAX_R - MIN_L), which is 10^5 * 2000 under given constraints. You program runs out of memory and gets runtime error. A more efficient algorithm is required. Here is a simple solution that uses only O(MAX_R) memory and O(N + MAX_R) time.
vector<int> start(MAX_R + 1);
vector<int> end(MAX_R + 1);
int n;
cin >> n;
for (int i = 0; i < n; i++) {
int low;
int high;
cin >> low >> high;
start[low]++;
end[high]++;
}
int res = 0;
int sum = 0;
for (int pos = 0; pos <= MAX_R; pos++) {
if (sum == 0 && start[pos] > 0)
res++;
sum += start[pos] - end[pos];
}
cout << res << endl;
There is no need for bfs or any other graph algorithms in this problem.
You could fix your original solution by avoiding multiple edges in the graph(there is no need to create an edge from i to i + 1 if it already exists, but I am not sure if your original solution is correct).

Answer 2

Looks like you should start by changing to vector<pair<int,int>> v;. Then to populate v you should use:
scanf("%d", &l);scanf("%d", &r);
v.push_back(make_pair(l, r);
Then your function should become something like:
int solve(){
vector<pair<int, int>> results;
for(auto& vIndex : v){
auto resultIndex = find_if(results.begin(), results.end(), [vIndex](const pair<int, int>& i){return vIndex.first >= i.first && vIndex.first <= i.second || vIndex.second >= i.first && vIndex.second <= i.second;});
if(resultIndex == results.end()){
results.push_back(vIndex);
}else{
resultIndex->first = min(vIndex.first, resultIndex->first);
resultIndex->second = max(vIndex.second, resultIndex->second);
}
}
return results.size();
}
You can see this in action here: http://ideone.com/MDQBOr Just hard code your desired inputs into v.