Problem:
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Connected Components

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 $n$ and $m$ are given. $n$ is the number of users in the SNS and $m$ is the number of relations in the SNS. The users in the SNS are identified by IDs $0, 1, ..., n-1$.

In the following $m$ lines, the relations are given. Each relation is given by two integers $s$ and $t$ that represents $s$ and $t$ are friends (and reachable each other).

In the next line, the number of queries $q$ is given. In the following $q$ lines, $q$ queries are given respectively. Each query consists of two integers $s$ and $t$ separated by a space character.

Output

For each query, print "yes" if $t$ is reachable from $s$ through the social network, "no" otherwise.

Constraints

$2 \leq n \leq 100,000$

$0 \leq m \leq 100,000$

$1 \leq q \leq 10,000$

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

Solution in C++14 (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int NIL=-1;
const int MAX=100000;

int n, color[MAX];
vector<int> G[MAX];

void dfs(int v, int id){
    stack<int> s;
    s.push(v);
    color[v]=id;

    while(!s.empty()){
        int u=s.top(); s.pop();
        for(int i=0; i<G[u].size(); i++){
            int v1=G[u][i];
            if(color[v1]==NIL){
                color[v1]=id;
                s.push(v1);
            }    
        }
    }
}

void assignColor(){
    int id=1;

    for(int i=0; i<n; i++) color[i]=NIL;
    for(int i=0; i<n; i++){
        if(color[i]==NIL){
            dfs(i,id); id++;
        }
    }
}

int main(void){
    int m; cin >> n >> m;

    for(int i=0; i<m; i++){
        int s,t; scanf("%d %d", &s, &t);
        G[s].push_back(t);
        G[t].push_back(s);
    }

    assignColor();

    int q; cin >> q;
    for(int i=0; i<q; i++){
        int s,t; scanf("%d %d", &s, &t);
        if(color[s]==color[t]) printf("yes\n");
        else printf("no\n");
    }

    return 0;
}