How Many Tables HDU

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

大致题意：一个人过生日，他请了n个朋友来，给他们安排桌子吃东西，有m组朋友是互相认识的，当某个人和这张桌子上所有人都不认识时，他就只能坐到新的桌子上，问最少需要多少张桌子。

思路：简单的并查集套模板，有多少个集合即最少需要多少张桌子。

代码如下

#include<iostream>#include<algorithm>#include<string>#include<cstdio>#include<cstring>#include<queue>#include<vector>#include<cstring>using namespace std;const int maxn=1005;int pre[maxn];int total;int n,m;int find(int x){   int r=x;   while (pre[r]!=r)   r=pre[r];   int i=x; int j;   while(i!=r)   {       j=pre[i];       pre[i]=r;       i=j;   }   return r;}void join(int x,int y){    int f1=find(x);    int f2=find(y);    if(f1!=f2)    {        pre[f2]=f1;        total--;    }}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&m);        total=n;        for(int i=1;i<=n;i++)        pre[i]=i;        for(int i=1;i<=m;i++)        {            int x,y;            scanf("%d%d",&x,&y);            join(x,y);        }        printf("%d\n",total);     }   return 0;}