luogu P2419 [USACO08JAN]牛大赛Cow Contest

题目描述

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.

FJ的N(1 <= N <= 100)头奶牛们最近参加了场程序设计竞赛:)。在赛场上，奶牛们按1..N依次编号。每头奶牛的编程能力不尽相同，并且没有哪两头奶牛的水平不相上下，也就是说，奶牛们的编程能力有明确的排名。 整个比赛被分成了若干轮，每一轮是两头指定编号的奶牛的对决。如果编号为A的奶牛的编程能力强于编号为B的奶牛(1 <= A <= N; 1 <= B <= N; A != B) ，那么她们的对决中，编号为A的奶牛总是能胜出。 FJ想知道奶牛们编程能力的具体排名，于是他找来了奶牛们所有 M(1 <= M <= 4,500)轮比赛的结果，希望你能根据这些信息，推断出尽可能多的奶牛的编程能力排名。比赛结果保证不会自相矛盾。

输入格式

第1行: 2个用空格隔开的整数：N 和 M

第2..M+1行: 每行为2个用空格隔开的整数A、B，描述了参加某一轮比赛的奶 牛的编号，以及结果（编号为A，即为每行的第一个数的奶牛为 胜者）

输出格式

第1行: 输出1个整数，表示排名可以确定的奶牛的数目

#include<iostream>
#include<cstdio>
using namespace std;
int n,m,f[101][101],ans;
int main(){
    cin>>n>>m;
    for(int i=1,a,b;i<=m;i++){
        scanf("%d%d",&a,&b);
        f[a][b]=1;
    }
    for(int k=1;k<=n;k++)
    for(int i=1;i<=n;i++)
    for(int j=1;j<=n;j++)
    f[i][j]=f[i][j]|(f[i][k]&f[k][j]);
    for(int i=1;i<=n;i++){
        int gg=1;
        for(int j=1;j<=n;j++)
        if(i==j)continue;
        else gg=gg&(f[i][j]|f[j][i]);
        ans+=gg;
    }
    printf("%d\n",ans);
}

标签：USACO08JAN,Cow,cow,Contest,int,results,cows,skill,rounds
来源： https://www.cnblogs.com/naruto-mzx/p/11862879.html