POJ

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题目连接：http://poj.org/problem?id=3660

题目描述

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.

有n(1<=n<=100)个学生参加编程比赛。

给出m条实力信息。(1<=M<=4500)

其中每一条的格式为 A B （1<=A<=N,1<=B<=N,A!=B) 意思是A的实力比B强。

如果A比B强且B比C强，那么A一定比C强。

问最后有多少名学生可以确定他的排名。

保证输入信息不存在矛盾

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

第一行n和m。

以下m行 A B 表示A实力比B强。

Output

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

输出答案

Sample Input

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

Sample Output

2

解题思路

对每给的一个胜负关系连一条边，最后跑一次Floyd，然后判断一头牛所确定的关系是否是n-1次，若是，则这头牛的排名可以确定
简单说就是这头牛可以和另一只不管怎样如果能连接到，就说明时是可以确定的

AC代码

#include<iostream>#include<cstring>using namespace std;int dis[110][110];int main () {    int n, m;    int a, b;    while(~scanf("%d %d", &n, &m)) {        memset(dis, 0, sizeof(dis));        for(int i = 1; i <= m; i++) {            cin >> a >> b;            dis[a][b] = 1;        }        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 ans = 0;        for(int i = 1; i <= n; i++) {            int sum = 0;            for(int j =1; j <= n; j++) {                if(dis[i][j] || dis[j][i])                    sum++;            }            if(sum == n-1)                 ans++;        }        printf("%d\n", ans);    }    return 0;}