hdu 1213 How Many Tables

How Many Tables

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8018    Accepted Submission(s): 3892

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
 
 
这其实就是并查集的简单题目，看可以构建多少个数而已
在判断有几个桌子的时候，我们一般的都是在合并完成后扫描整个父亲数组，发现几个father[i]==i,就判断有多少个树，从而需要多少个桌子
 
AC代码：
#include <iostream>#include <cstdio>using namespace std;int father[1005];int getfather(int n){    while(n!=father[n])    n=father[n];    return n;}int main(){    int n,m,t,a,b;    cin>>t;    while(t--)    {        cin>>n>>m;        for(int i=1;i<=n;i++)        father[i]=i;        for(int i=1;i<=m;i++)        {            cin>>a>>b;            int roota=getfather(a);            int rootb=getfather(b);            if(roota!=rootb)            father[rootb]=roota;        }        int cnt=0;        for(int i=1;i<=n;i++)        if(father[i]==i) cnt++;        cout<<cnt<<endl;    }    return 0;}