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
2
5 3
1 2
2 3
4 5
5 1
2 5
Sample Output
2
4
题意:
今天是伊格那丢的生日。他邀请了很多朋友。现在该吃晚饭了。伊格那丢想知道他至少需要多少张桌子。你必须注意到,不是所有的朋友都互相认识,所有的朋友都不想和陌生人待在一起。
这个问题的一个重要规则是如果我告诉你A认识B, B认识C,这意味着A, B, C互相认识,所以它们可以在一个表中。
例如:如果我告诉你A知道B, B知道C, D知道E,那么A、B、C可以留在一个表中,而D、E必须留在另一个表中。伊格那丢至少需要两张表。
输入
输入以一个整数T(1<=T<=25)开始,它表示测试用例的数量。然后是测试用例。每个测试用例都以两个整数N和M开始(1<=N,M<=1000)。N表示朋友的数量,从1到N标记朋友,然后M行跟随。每一行由两个整数A和B(A!=B)组成,这意味着朋友A和朋友B相互认识。两种情况之间会有一条空行。
输出
对于每个测试用例,只需输出Ignatius至少需要多少个表。不要打印任何空格。
解析:一共有n个联通分量,每一次有效的合并都会-1
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
const int N=1e5+100;
int fa[N],n,m,a,b;
int ans[N];
int t;
int find(int x)
{
if(x!=fa[x]) return fa[x]=find(fa[x]);
return fa[x];
}
int build(int a,int b)
{
int x=find(a);
int y=find(b);
if(x!=y)
{
fa[x]=y;
return 1;
}
return 0;
}
int main()
{
cin>>t;
while(t--)
{
cin>>n>>m;
for(int i=1;i<=n;i++) fa[i]=i;
int res=n;
while(m--)
{
cin>>a>>b;
res-=build(a,b);
}
cout<<res<<endl;
}
}