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 (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
 
int n, m;
 
void merge(int *u, int *net, int oldnet, int newnet);
 
int main(int argc, char *argv[])
{
  int i, s, t, p, q;
  int *u;
  int *net;
 
  scanf("%d", &n);
  scanf("%d", &m);
  u = (int *)malloc(sizeof(int) * n);
  net = (int *)malloc(sizeof(int) * (m + 1));
  for (i = 0; i < n; i++) u[i] = -1;
  for (i = 0; i <= m; i++) net[i] = 0;
 
  p = 1;
 
  for (i = 0; i < m; i++) {
    scanf("%d", &s);
    scanf("%d", &t);
    if (s != t) {
      if (u[s] == -1 && u[t] == -1) {
    u[s] = p;
    u[t] = p;
    net[p] += 2;
    p++;
      } else {
    if (u[s] != -1 && u[t] == -1) {
      u[t] = u[s];
      net[u[s]]++;
    } else if (u[s] == -1 && u[t] != -1) {
      u[s] = u[t];
      net[u[t]]++;
    } else {
      if (u[s] != u[t]) {
        if (net[u[s]] > net[u[t]]) merge(u, net, u[t], u[s]);
        else merge(u, net, u[s], u[t]);
      }
    }
      }
    }
  }
 
  scanf("%d", &q);
 
  for (i = 0; i < q; i++) {
    scanf("%d", &s);
    scanf("%d", &t);
 
    if (s == t) printf("yes\n");
    else {
      if (u[s] == u[t] && u[s] != -1 && u[t] != -1) printf("yes\n");
      else printf("no\n");
    }
  }
 
  return 0;
}
 
void merge(int *u, int *net, int oldnet, int newnet)
{
  int i;
   
  for (i = 0; i < n; i++) {
    if (u[i] == oldnet) u[i] = newnet;
  }
 
  net[newnet] += net[oldnet];
  net[oldnet] = 0;
}