poj3660Cow Contest（Floyd闭包）

题意：给出一些牛的强弱关系，问有多少只牛能确定其排名

思路：把牛看成顶点，A强于B就在A和B连一条边，然后用Floyd跑一遍传递闭包，如果一只牛赢的次数和输的次数加起来等于n-1，那么就可以确定了

吐槽：一开始以为只要某个顶点连入的边数等于n-1就可以了，（怎么会这么简单）后来WA了一次之后发现如果A赢B，B赢C，就算A不和C打也能确定A能赢C

#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define maxn 505
#define LL long long
int cas=1,T;
int d[maxn][maxn];
const int INF = 1<<29;
int n,m;
int main()
{
scanf("%d%d",&n,&m);
for (int i = 0;i<m;i++)
{
int a,b;
scanf("%d%d",&a,&b);
d[a][b]=1;

}
for (int k = 1;k<=n;k++)
for (int i = 1;i<=n;i++)
for (int j = 1;j<=n;j++)
d[i][j]=d[i][j] || (d[i][k] && d[k][j]);
int ans = 0;
for (int i = 1;i<=n;i++)
{
int anss = 0;
for (int j = 1;j<=n;j++)
{
anss+=d[i][j]+d[j][i];
}
if (anss==n-1)
ans++;
}
printf("%d\n",ans);
//freopen("in","r",stdin);
//scanf("%d",&T);
//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
return 0;
}

题目

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 5
4 3
4 2
3 2
1 2
2 5

Sample Output

2