| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 12848 | Accepted: 7149 |
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
Source
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題意：
牛在比賽，問知道兩頭牛的關係。最終可以確定多少個牛的名次
思路：
只要找到某隻牛與其他所有牛的關係則其名次一定可以確定
代碼：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
int dis[105][105];
int n,m;
void floyd()
{
for(int k=1;k<=n;k++)
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
//i 可直接或間接打敗 j
if(dis[i][j] || (dis[i][k]&&dis[k][j]))
dis[i][j]=1;
}
int main()
{
scanf("%d%d",&n,&m);
memset(dis,0,sizeof(dis));
for(int i=0;i<m;i++)
{
int a,b;
scanf("%d%d",&a,&b);
dis[a][b]=1;//不能雙向
}
floyd();
int sum=0;
for(int i=1;i<=n;i++)
{
int t=0;
for(int j=1;j<=n;j++)
{
if(dis[i][j] || dis[j][i])t++;
}
//只要確定了自己和其他所有的關係，則名次一定確定
if(t==n-1)sum++;
}
printf("%d\n",sum);
return 0;
}