POJ 3660 Cow Contest (Floyd 傳遞閉包)

POJ 3660 Cow Contest (Floyd 傳遞閉包)

Cow Contest Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7443 Accepted: 4133

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 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), 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

題目鏈接：http://poj.org/problem?id=3660

題目大意：n個人，m組關系，關系a b表示a排名在b前，現在要求有多少個人的排名已經確定了

題目分析：用Floyd求傳遞閉包，a在b前，b在c前，則a在c前，跑完Floyd每個點和別的點的位置關系若都已經確定，則其名次已經確定

#include 
#include 
int const MAX = 105;

bool ok[MAX][MAX];
int n, m;

void Floyd()
{
    for(int k = 1; k <= n; k++)
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                if(ok[i][k] && ok[k][j])
                    ok[i][j] = true;
}

int main()
{
    scanf("%d %d", &n, &m);
    memset(ok, false, sizeof(ok));
    for(int i = 0; i < m; i++)
    {
        int a, b;
        scanf("%d %d", &a, &b);
        ok[a][b] = true;
    }
    Floyd();
    int ans = 0;
    for(int i = 1; i <= n; i++)
    {
        bool flag = true;
        for(int j = 1; j <= n; j++)
        {
            if(i == j)
                continue;
            if(!ok[i][j] && !ok[j][i])
            {
                flag = false;
                break;
            }
        }
        if(flag)
            ans ++;
    }
    printf("%d\n", ans);
}