杭电1213

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

此题算法：并查集 详见———>>百度百科
首先初始化x[i]=i;
将互相认识的合并成集合。（本质就是连通图，将互相认识的连通，最后判断是否在一个连通图中）

#include<stdio.h>int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,m,x[1000],k,g,num=0,a[1005],b[1005];        scanf("%d%d",&n,&m);        for(int i=0; i<=n; i++)            x[i]=i;        for(int i=0; i<m; i++)            scanf("%d%d",&a[i],&b[i]);        for(int i=0; i<m; i++)        {            for(k=a[i]; x[k]!=k; k=x[k]);            for(g=b[i]; x[g]!=g; g=x[g]);            if(k!=g)                x[g]=k;        }        for(int i=1; i<=n; i++)        {            if(x[i]==i)                num++;            //printf("x[%d]=%d\n",i,x[i]);        }        printf("%d\n",num);    }}