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POJ3660 Cow Contest(floyd传递闭包)

2018 年 08 月 23 日 • 阅读: 1311 • 图论阅读设置

Cow Contest

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 15880 Accepted: 8852

Description

N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.

The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ AN; 1 ≤ BN; AB), then cow A will always beat cow B.

Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B

Output

* Line 1: A single integer representing the number of cows whose ranks can be determined

Sample Input

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

Sample Output

2

Source

USACO 2008 January Silver


链接

http://poj.org/problem?id=3660

题意

n头奶牛,每头奶牛都有一个唯一的技能排名,如果奶牛A能打败奶牛B,说明A的排名比B高,具有传递性,现在给定m场比赛信息,问有多少头奶牛能确定技能排名。

题解

Floyd传递闭包。跑一遍floyd就得到了任意两头奶牛之间是否能确定关系,如果一头奶牛与其余的n-1头奶牛都能确定关系,那么它的排名就一定能确定。

代码

StatusAccepted
Time32ms
Memory700kB
Length1053
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 1010;
int G[maxn][maxn];
int n, m;

void floyd() // floyd求传递闭包
{
    for (int k = 1; k <= n; ++k)
    {
        for (int i = 1; i <= n; ++i)
        {
            for (int j = 1; j <= n; ++j)
            {
                if (G[i][k] && G[k][j]) // i->k->j
                    G[i][j] = 1;
            }
        }
    }
}

int main()
{
    int u, v;
    while (scanf("%d%d", &n, &m) != EOF)
    {
        memset(G, 0, sizeof(G));
        for (int i = 0; i < m; ++i)
        {
            scanf("%d%d", &u, &v);
            G[u][v] = 1;
        }
        floyd();
        int cnt, ans = 0;
        for (int i = 1; i <= n; ++i)
        {
            cnt = 0;
            for (int j = 1; j <= n; ++j)
            {
                if (G[i][j] || G[j][i]) // i和j的关系确定
                    cnt++;
            }
            if (cnt == n - 1) // 该点和别的n-1个点的关系都确定
                ++ans;
        }
        printf("%d\n", ans);
    }
    return 0;
}

一道大水题,了解一下floyd传递闭包。


还有一道题,HDU1704 RANK ,为啥上面的代码蜜汁超时了呢!和下面的也差不了多少。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 510;
int G[maxn][maxn];
int n, m;

void floyd() // floyd求传递闭包
{
    for (int k = 1; k <= n; ++k)
    {
        for (int i = 1; i <= n; ++i)
        {
            if (G[i][k])
            {
                for (int j = 1; j <= n; ++j)
                {
                    if (G[k][j]) // i->k->j
                        G[i][j] = 1;
                }
            }
        }
    }
}

int main()
{
    int T, u, v;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d%d", &n, &m);
        memset(G, 0, sizeof(G));
        for (int i = 0; i < m; ++i)
        {
            scanf("%d%d", &u, &v);
            G[u][v] = 1;
        }
        floyd();
        int ans = 0;
        for (int i = 1; i <= n; ++i)
        {
            for (int j = i + 1; j <= n; ++j)
            {
                if (!(G[i][j] || G[j][i])) // i和j的关系不确定
                    ans++;
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}

The end.
2018-08-23 星期四