HDU1213How Many Tables

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

Sample Output

2
4

解法分析及AC代码

此题为并查集的模板题,只要找到不同集合的个数再计数即可，具体见代码及注释。

#include<bits/stdc++.h>
using namespace std;
#define maxn 1005
int t,n,m,total;
int a[maxn],b[maxn];//数组a存入的是每个人所属的集合的名称int find(int x) //这个函数用来找到自己所属的集合，返回的是自己所属集合的名称
{return x==a[x]?x:a[x]=find(a[x]);//使用路径压缩
}
void sett(int n)//初始化将一开始每个人所属的集合的名称定义成自己的序号
{for(int i=1;i<=n;i++){a[i]=i;}
}void sum(int x,int y)//此函数用来合并两个朋友的所属集合，使它们在一个集合里面
{if(x!=y) a[x]=y;
}
int main()
{scanf("%d",&t);while(t--){memset(b,0,sizeof(b));total=0;scanf("%d%d",&n,&m);sett(n);for(int i=1;i<=m;i++){int x,y;scanf("%d%d",&x,&y);sum(find(x),find(y));}for(int i=1;i<=n;i++){//在同一个集合中的人共用一张桌子，一个集合只被计数一次,下面是计数一次的方法if(b[find(a[i])]++==0) total++;}printf("%d\n",total);}
}

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HDU1213How Many Tables

解法分析及AC代码

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