hdu 1213 求树的个数

Problem Description
Today is Ignatius’ birthday. He invites a lot of friends. Now it’s dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want to stay with strangers.

One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table.

For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least.

Input
The input starts with an integer T(1<=T<=25) which indicate the number of test cases. Then T test cases follow. Each test case starts with two integers N and M(1<=N,M<=1000). N indicates the number of friends, the friends are marked from 1 to N. Then M lines follow. Each line consists of two integers A and B(A!=B), that means friend A and friend B know each other. There will be a blank line between two cases.

Output
For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.

Sample Input
2
5 3
1 2
2 3
4 5

5 1
2 5

Sample Output
2
4

题意：

并查集模板题。不知道为什么刚开始一直内存超出。怀疑OJ有毒。
其实就是求树的个数！父节点等于本身的点是根节点！

代码：

#include <iostream>
#include <cstdio>
using namespace std;
const int maxn = 1002;

int par[maxn];
int n,m;
void init()
{
    for(int i=0;i<=n;i++)
    {
        par[i]=i;
    }
}

int solve(int x)
{
   if(par[x]==x)
   {
       return x;
   }
   else
   {
       return par[x] = solve(par[x]);
   }
}

void unite(int x,int y)
{
    int pa = solve(x);
    int pb = solve(y);
    if(pa==pb) return;
    par[pa]=pb;
}

int main()
{
    int T;

    int x,y;
    scanf("%d",&T);

    while(T--)
    {
       scanf("%d%d",&n,&m);
       init();
       for(int i=0;i<m;i++)
       {
           scanf("%d%d",&x,&y);
           unite(x,y);
       }

       int ans=0;

       for(int i=1;i<=n;i++)
       {
            if(par[i]==i)
                ans++;
       }

       printf("%d\n",ans);

     }

    return 0;
}