最短路练习 8/poj/3660/Cow Contest

题目链接：http://poj.org/problem?id=3660

Cow Contest

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11308 Accepted: 6280

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 54 34 23 21 22 5

Sample Output

2

题意：有n头牛，有m组关系，这m组关系不一定能得出所有的牛的排名，可能会有一些关系不确定，最多可以确定几头牛的排名。

思路：用Floyd：a>b,b>c 能得到a>c;     a>b,c>b得不出a和c；  所以关系是单向的；

有关系的为1，没关系的为0；最后确定一头牛的排名确定不确定，看 他若与其它牛的关系都确定，排名则确定。

最后统计数量。

可惜了，我到最后统计数量的时候没想到这个方法。

AC代码：

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<vector>#include<map>#include<string>#define LL long long#define eps 1e-8using namespace std;const int mod = 1e7+7;const int inf = 0x3f3f3f3f;const int maxn = 1e6 +10;int n,m,a,b;int ma[105][105];int main(){    while(~scanf("%d%d",&n,&m))    {        memset(ma,0,sizeof(ma));        for(int i=0;i<m;i++)        {            scanf("%d%d",&a,&b);            ma[a][b]=1;        }        for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++)            for(int k=1;k<=n;k++)                 if(ma[j][i]&&ma[i][k])                     ma[j][k]=1;        int sum,ans=0;        for(int i=1;i<=n;i++)        {            sum=0;            for(int j=1;j<=n;j++)                if(ma[i][j]||ma[j][i])                    sum++;            if(sum==n-1)                ans++;        }        printf("%d\n",ans);    }}