HDU-1213 How Many Tables（并查集 模板）

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描述

传送门：HDU-1213 How 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.

输入描述

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.

输出描述

For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.

示例

输入

1
2
3
4
5
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2
5 3
1 2
2 3
4 5

5 1
2 5

输出

1
2
2
4

题解

题目大意

要宴请一些客人，但是客人彼此之间并不是完全互相认识的，规定如果A认识B，且B认识C，则认为A,B,C互相认识，可以安排在同一桌，求需要安排的桌数。

思路

并查集模板题，水题。

代码

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#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5;
int Fa[MAXN];

int Find(int x){
    if(Fa[x] == x) return x;
    else return Fa[x] = Find(Fa[x]);
}

void unite(int x, int y){
    x = Find(x);
    y = Find(y);
    if(x == y) return;
    else{
        Fa[x] = y;
    }
}

int main(){
    int t, n, m, x, y;
    cin >>t;
    while(t--){
        cin >> n >> m;
        for(int i = 1; i <= n; i++) Fa[i] = i;
        while(m--){
            cin >> x >> y;
            unite(x, y);
        }
        int ans = 0;
        for(int i = 1; i <= n; i++){
            if(Fa[i] == i) ans++;
        }
        printf("%d\n", ans);
    }
}