[USACO08JAN] Cow Contest

题目背景

[Usaco2008 Jan]

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。

输入输出格式

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

 

输入输出样例

输入样例#1： 

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

输出样例#1： 

2

说明

输出说明:

编号为2的奶牛输给了编号为1、3、4的奶牛，也就是说她的水平比这3头奶

牛都差。而编号为5的奶牛又输在了她的手下，也就是说，她的水平比编号为5的

奶牛强一些。于是，编号为2的奶牛的排名必然为第4，编号为5的奶牛的水平必

然最差。其他3头奶牛的排名仍无法确定。

显然用Floyed传递闭包

code:

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>

using namespace std;

const int N = 110;

int n, m, ans, vs[N][N];

inline int read() {
    int num = 0, f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') f = -1;
        ch = getchar();
    }
    while (isdigit(ch)) {
        num = num * 10 + ch - '0';
        ch = getchar();
    }
    return num * f;
}

int main() {
    memset(vs, 0, sizeof(vs));
    
    n = read(); m = read();
    for (int i = 1; i <= m; i++) {
        int a, b;
        a = read(); b = read();
        vs[a][b] = 1;
    }
    
    //Floyed传递闭包 
    for (int k = 1; k <= n; k++)
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                vs[i][j] |= (vs[i][k] & vs[k][j]);  
                //表示i能否走到j，即要么一开始i能到j,要么i能到k，k再能到j。
                
    for (int i = 1; i <= n; i++) {
        int temp = 0;
        
        for (int j = 1; j <= n; j++) 
            if (vs[i][j] || vs[j][i]) temp++;
    
        if (temp == n - 1) ans++; 
        //如果被打败和打败的总数为n-1，即被其他所有牛访问过或间接访问过，那么它一定是可以确定的 
    }
    
    printf("%d", ans);
    
    return 0;
}