滴答滴答---題目連結 

Write a program which reads relations in a SNS (Social Network Service), and judges that given pairs of users are reachable each other through the network.

Input

In the first line, two integer nn and mm are given. nn is the number of users in the SNS and mm is the number of relations in the SNS. The users in the SNS are identified by IDs 0,1,...,n−10,1,...,n−1.

In the following mm lines, the relations are given. Each relation is given by two integers ss and tt that represents ss and tt are friends (and reachable each other).

In the next line, the number of queries qq is given. In the following qq lines, qq queries are given respectively. Each query consists of two integers ss and tt separated by a space character.

Output

For each query, print "yes" if tt is reachable from ss through the social network, "no" otherwise.

Constraints

• 2≤n≤100,0002≤n≤100,000
• 0≤m≤100,0000≤m≤100,000
• 1≤q≤10,0001≤q≤10,000

Sample Input

10 9
0 1
0 2
3 4
5 7
5 6
6 7
6 8
7 8
8 9
3
0 1
5 9
1 3

Sample Output

yes
yes
no

#include <bits/stdc++.h>
using namespace std;
int n;
const int maxn=100001;
vector<int>g[maxn];
const int NLL=-1;
int color[maxn];
void dfs(int r,int c)
{
    stack<int>s;
    s.push(r);
    color[r]=c;
    while(!s.empty())
    {
        int u=s.top();
        s.pop();
        for(int i=0; i<(int)g[u].size(); i++)//搜尋與u頂點相連的點v
        {
            int v=g[u][i];
            if(color[v]==NLL)
            {
                color[v]=c;
                s.push(v);
            }
        }
    }
}
void assigncolor()
{
    int id=1;
    for(int i=0; i<n; i++)
        color[i]=NLL;
    for(int i=0; i<n; i++)
        if(color[i]==NLL)
            dfs(i,id++);
}
int main()
{
    int m,q;
    cin>>n>>m;
    for(int i=0; i<m; i++)
    {
        int x,y;
        cin>>x>>y;
        g[x].push_back(y);
        g[y].push_back(x);

    }
    assigncolor();
    cin>>q;
    while(q--)
    {
        int x,y;
        cin>>x>>y;
        if(color[x]==color[y])
            printf("yes\n");
        else printf("no\n");
    }
    return 0;
}