How Many Tables(并查集)

最新推荐文章于 2021-08-08 13:18:43 发布

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本文介绍了一个关于朋友分组的问题，使用并查集算法解决如何最少分配餐桌数量的问题。通过输入朋友之间的熟悉关系，确定最少需要多少张桌子才能让每个熟悉的朋友都能坐在同一桌。

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How Many Tables

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23308 Accepted Submission(s): 11604

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

Sample Output

2
4

Author

Ignatius.L

Source

杭电ACM省赛集训队选拔赛之热身赛

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#include <iostream>
#include <cstdio>
#include <set>

using namespace std;

const int maxn = 1000+100;

int high[maxn];
int par[maxn];

void init(int n)
{
    for (int i=1; i<=n; i++) //啊啊啊，终于找到你，1~n过，0~n-1就WA
    {
        high[i] = 0;
        par[i] = i;
    }
}

int findr(int x)
{
    if (par[x] == x)
    {
        return x;
    }
    else
    {
        return par[x] = findr(par[x]);
    }
}

void uni(int x, int y)
{
    x = findr(x);
    y = findr(y);
    if (x == y)
    {
        return ;
    }
    if (high[x] > high[y])
    {
        par[y] = x;
    }
    else
    {
        par[x] = y;
        if (high[x] == high[y])
        {
            high[y]++;
        }
    }
}

void ans(int n)
{
    set<int> s;
    for (int i=1; i<=n; i++) //啊啊啊，终于找到你，1~n过，0~n-1就WA
    {
        s.insert(findr(i));
        cout << findr(i) << endl;
    }
    cout << s.size() << endl;
}

int main()
{
    int t;
    int n, m;
    int x, y;
    scanf("%d", &t);
    while( t-- )
    {
        scanf("%d%d", &n, &m);
        init(n);
        for (int i=0; i<m; i++)
        {
            scanf("%d%d", &x, &y);
            uni(x, y);
        }
        ans(n);
    }
    return 0;
}

哎。别人都在写新的任务，我就返回来为了一道麻烦的cccc写了会简单的并查集，哎，做题做的太少啦~~