Cow Contest
- 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
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* 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
-
题意:给你n头牛,每头牛都有自己的kills值,和其他牛的都不同,然后kills大的可以赢小的。现在给出n头牛的比赛结果,问你有几头牛可以确定排名。
如果A>B,B>C,那么A>C,所以这道题我们要用到floyd的闭包传递。
AC代码:
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
#define INF 0x3f3f3f3f
using namespace std;
int map[101][101];
int main()
{
int n, m;
while(~scanf("%d%d",&n,&m))
{
if(n==0&&m==0)
break;
memset(map, 0, sizeof(map));
for(int i = 1; i <= m; i++)
{
int x, y;
scanf("%d%d",&x,&y);
map[x][y] = 1;
}
for(int k = 1; k <= n; k++)
{
for(int i = 1; i <= n; i++)
{
for(int j = 1; j <= n; j++)
{
if(map[i][k]&&map[k][j])
{
map[i][j] = 1;
}
}
}
}
int c1 = 0, c2 = 0;
int ans = 0;
for(int i = 1; i <= n; i++)
{
c1 = 0, c2 = 0;
for(int j = 1; j <= n; j++)
{
if(map[i][j])
c1++;
if(map[j][i])
c2++;
}
if(c1+c2==n-1)
ans++;
}
printf("%d\n",ans);
}
return 0;
}