1213 How Many Tables

How Many Tables

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 18167    Accepted Submission(s): 8944

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

25 31 22 34 55 12 5

 

Sample Output

24

题意，有个人要请客，请了很多朋友，规定，相互是朋友的就坐在一起，否则就需要另一个桌子.....

给出第一个数，代表几组测试数据，每组前两个数，表示一共多少人（编号从1~n），第二个数表示，这些人有多少关系，

下面几行就是这些关系对应的哪些人，也就是哪两个人认识.......

分析：

这个相当于统计一共有多少个集合的问题，用并查集，只要统计出有多少父节点（就是多少堆人），也就统计出来了需要多少桌子......

#include<stdio.h>#include<string.h>int per[1005];void init(int n)//初始化{    for(int i=1;i<=n;++i)    {        per[i]=i;    }}int find(int x){    int r=x;    while(per[r]!=r)//查找    {        r=per[r];    }    int i=x,j;    while(i!=r)//压缩路径    {        j=per[i];per[i]=r;i=j;    }    return r;}void join(int x, int y){    int fx=find(x),fy=find(y);    if(fx!=fy)//合并    {        per[fx]=fy;    }}int main(){    int t,n,m,i,a,b,c;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        init(n);c=0;        while(m--)        {            scanf("%d%d",&a,&b);            join(a,b);        }        for(i=1;i<=n;++i)        {            if(per[i]==i)//查找父节点.....            {                ++c;            }        }        printf("%d\n",c);    }    return 0;}