How Many Tables

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题目描述

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.

输入

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.

输出

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

样例输入

2
6 4
1 2
2 3
3 4
1 4

8 10
1 2
2 3
5 6
7 5
4 6
3 6
6 7
2 5
2 4
4 3

样例输出

3
2

题目类型

并查集

题意概括

判断几个人之间是否认识，认识的坐一张桌子，不认识的就需要做不同的桌子，问最少需要多少张桌子。

解题思路

利用并查集算法，随后输出一共有多少种集合。

代码如下：

#include<stdio.h>#include<string.h>#include<ctype.h>#include<math.h>int f[1100],n,m,k,sum;void init (){    int i;    for(i=1;i<=n;i++){        f[i]=i;    }    return ;}int getf(int v){    if(f[v]==v){        return v;    }else {        f[v]=getf(f[v]);        return f[v];    }}void marge (int v,int u){    int t1,t2;    t1=getf(v);    t2=getf(u);    if(t1!=t2){        f[t2]=t1;    }    return ;}int main (){    int i,x,y,j;    int t;    scanf("%d",&t);    for(j=0;j<t;j++){        memset(f,0,sizeof(f));        sum=0;        scanf("%d%d",&n,&m);        init();        for(i=1;i<=m;i++){            scanf("%d %d",&x,&y);            marge(x,y);        }        for(i=1;i<=n;i++){            if(f[i]==i)                sum++;        }        printf("%d\n",sum);    }}