Cow Contest(Floyd)

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Cow Contest
时间限制:1000 ms | 内存限制:65535 KB
难度:4
描述
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.

输入
* 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

There are multi test cases.The input is terminated by two zeros.The number of test cases is no more than 20.
输出
For every case:
* Line 1: A single integer representing the number of cows whose ranks can be determined
样例输入
5 5
4 3
4 2
3 2
1 2
2 5
0 0
样例输出
2
来源
USACO 2008 January Silver

分析:
如何确定一个数的等级呢?有n个数,如果一个数比x个数大,比y个数小,且x+y=n-1则它的位置就确定了。那如果确定有多少个数比他大或小?如果a>b则有边a->b,那么只要一个数a可达另一个数b则说明a一定大于b。我们用Floyd算法求出每个数之间的可达关系,即可求出大小关系。

AC代码:

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn=110;int E[maxn][maxn];int d[maxn];int main(){    int n,m,a,b;    while(scanf("%d%d",&n,&m)!=EOF)    {        if(n==0 && m==0) break;        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                if(i==j) E[i][j]=1;                else E[i][j]=0;            }        }        for(int i=0;i<m;i++)        {            scanf("%d%d",&a,&b);            E[a][b]=1;        }        for(int k=1;k<=n;k++)        {            for(int i=1;i<=n;i++)            {                for(int j=1;j<=n;j++)                {                    E[i][j]=E[i][j]|(E[i][k]&E[k][j]);                }            }        }        memset(d,0,sizeof(d));        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                if(i!=j && E[i][j])                {                    d[i]++;d[j]++;                }            }        }        int ans=0;        for(int i=1;i<=n;i++)            if(d[i]==n-1)            ans++;        printf("%d\n",ans);    }    return 0;}
0 0