Cow Contest

题目链接：

http://acm.hust.edu.cn/vjudge/contest/122685#problem/H

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.

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##Input

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* 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

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##Output

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

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##Sample Input

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5 5

4 3

4 2

3 2

1 2

2 5

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##Sample Output

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2

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##Hint

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</big>

<br/>

##题意：

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给出N个点，M个点对：

每条点对 A B 意味着A点的权值大于B点.

现在要对这些点进行权值排名，求有多少个点的排名能够确定.

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<br/>

##题解：

<big>

将样例画一遍就比较容易看出来：

若某点跟其他n-1个点都联通，则这个点的排名可以确定. 否则不能.

问题就转换为了求n个点之间的联通关系.

而floyd算法正好可以求任意两点的联通关系，只需要把求最短路时的松弛操作修改一下即可.

dis[i][j] = dis[i][j] || (dis[i][k] && dis[k][j]);

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<br/>

##代码：

``` cpp

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<algorithm>

#include<queue>

#define mid(a,b) ((a+b)>>1)

#define LL long long

#define maxn 110

#define inf 0x3f3f3f3f

#define IN freopen("in.txt","r",stdin);

using namespace std;

int n, m;

bool dis[maxn][maxn];

void floyd() {

    for(int k=1; k<=n; k++)

        for(int i=1; i<=n; i++)

            for(int j=1; j<=n; j++)

                dis[i][j] = dis[i][j] || (dis[i][k] && dis[k][j]);

}

int main(int argc, char const *argv[])

{

    //IN;

    while(scanf("%d %d", &n,&m) != EOF)

    {

        memset(dis, 0, sizeof(dis));

        for(int i=1; i<=n; i++) dis[i][i] = 1;

        for(int i=1; i<=m; i++) {

            int u,v; scanf("%d %d", &u,&v);

            dis[u][v] = 1;

        }

        floyd();

        int ans = 0;

        for(int i=1; i<=n; i++) {

            int cnt1=0, cnt2=0;

            for(int j=1; j<=n; j++) {

                if(i == j) continue;

                if(dis[i][j]) cnt1++;

                if(dis[j][i]) cnt2++;

            }

            if(cnt1+cnt2 == n-1) ans++;

        }

        printf("%d\n", ans);

    }

    return 0;

}

巴特西

POJ 3660 Cow Contest （floyd求联通关系）

Cow Contest

题目链接：

Description

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