HDU 1213 How Many Tables （并查集）

How Many Tables

题目链接：

http://acm.hust.edu.cn/vjudge/contest/123393#problem/C

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

2

5 3

1 2

2 3

4 5

5 1

2 5

Sample Output

2

4

题意：

n个人参加晚宴;

完全不认识的两个人不能被分配在同一餐桌;

认识具有传递性：A认识B B认识C，那么A和C也认识.

题解：

很明显的并查集模版题.

将认识两个人合并到同一集合；

最后统计有多少个不同的集合即可;

注意：部分编译器不允许变量名与关键字冲突；故注意next和rank这些关键字，以免CE.

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <vector>
#define LL long long
#define eps 1e-8
#define maxn 1100
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;

int fa[maxn];
int _rank[maxn];

void init_set() {
    for(int i=0; i<maxn; i++) {
        fa[i] = i;
        _rank[i] = 0;
    }
}

int find_set(int x) {
    return fa[x] = (x==fa[x]? x:find_set(fa[x]));
}

void unit_set(int x, int y) {
    x = find_set(x);
    y = find_set(y);
    if(_rank[x] < _rank[y]) swap(x, y);
    fa[y] = x;
    if(_rank[x] == _rank[y]) _rank[x]++;
}

int n, m;

int main(int argc, char const *argv[])
{
    //IN;

    int t;  cin >> t;
    while(t--)
    {
        cin >> n >> m;
        init_set();
        while(m--) {
            int x,y; scanf("%d %d", &x, &y);
            unit_set(x,y);
        }

        int cnt = 0;
        bool ans[maxn] = {0};
        for(int i=1; i<=n; i++) {
            ans[find_set(i)] = 1;
        }
        for(int i=1; i<=n; i++)
            if(ans[i]) cnt++;

        printf("%d\n", cnt);
    }

    return 0;
}