HDOJ1213. How many tables（并查集连通块计数）

How Many Tables

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

Total Submission(s): 33891    Accepted Submission(s): 16955

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

 

Author

Ignatius.L

 

Source

杭电ACM省赛集训队选拔赛之热身赛  

【分析】并查集连通块计数。类似畅通工程（HDOJ1232），属于同一朋友圈的朋友坐在一桌，也就是他们同在一个连通块。连通块的个数即至少需要的桌子数。

#include <stdio.h>const int maxn=1010;int T;int N,M,A,B;int ans;struct friends{    int pre;    int rank;} f[maxn];void Init(int n){    int i;    ans=n;    for(i=1;i<=n;i++)    {        f[i].pre=i;        f[i].rank=0;    }}int Findpre(int n){    if(f[n].pre==n)        return n;    return f[n].pre=Findpre(f[n].pre);}void Union(int x,int y){    x=Findpre(x);    y=Findpre(y);    if(x==y)        return;    ans--;    if(f[x].rank<f[y].rank)        f[x].pre=y;    else    {        f[y].pre=x;        if(f[x].rank==f[y].rank)            f[x].rank++;    }}int main(){    int i;    scanf("%d",&T);    while(T--)    {        scanf("%d %d",&N,&M);        Init(N);        for(i=0;i<M;i++)        {            scanf("%d %d",&A,&B);            Union(A,B);        }        printf("%d\n",ans);    }    return 0;}