hdu1213 How Many Tables-优快云博客

本篇介绍了一个关于朋友间相识关系的问题，通过输入的朋友数量及相互认识的关系，使用并查集算法来计算最少需要多少张桌子进行合理安排。

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How Many Tables
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19795 Accepted Submission(s): 9856

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

2
5 3
1 2
2 3
4 5

5 1
2 5

Sample Output

2
4

 与hdu1232大同小异，求连通分量个数。

#include<iostream>
#include<stdio.h>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstring>
#include<string.h>
#include<map>
#include<queue>
#include<list>
#include<stack>
#include<cctype>
#include<fstream>
#include<sstream>
#include<iomanip>
#include<set>
#include<vector>
#include<cstdlib>
#include<time.h>
using namespace std;

int p[1005];
int find(int x)
{
    if (x != p[x])
        p[x] = find(p[x]);
    return p[x];
}
void unio(int x, int y)
{
    int px = find(x);
    int py = find(y);
    if (px == py)
        return;
    p[px] = py;
}
void make(int x)
{
    for (int i = 1; i <= x; i++)
        p[i] = i;
}
int main()
{
    int n, m;
    int t;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &n);
        scanf("%d", &m);
        int a, b;
        make(n);
        for (int i = 1; i<= m; i++)
        {
            scanf("%d%d", &a, &b);
            if (find(a) != find(b))
                unio(a, b);
        }
        int s = 0;
        for (int i = 1; i <= n; i++)
        {
            if (p[i] == i)
                s++;
        }
        printf("%d\n", s);
    }
    return 0;
}