题目
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
题目大意
给定n头牛，每头牛有一个1到n的值，且任意两头牛的值不同，若知道m个牛的值的次序，问有多少头牛的值可以确定
题解
若st[i][j]为0表示i,j关系不确定，为1表示第i头牛的值大于第j头，为-1表示小于第j头牛，则第i头牛的值能确定的条件为对于每个k≠i，st[i][k]≠0.
代码
#include <iostream>
#include <cstring>
using namespace std;
const int N = 110, M = 4510;
int st[N][N];
int main()
{
memset(st,0,sizeof st);
int n,m;
cin >> n >> m;
for(int i = 1; i <= m; i ++)
{
int u, v;
cin >> u >> v;
st[u][v] = 1;
st[v][u] = -1;
for(int j = 1; j <= n; j ++)
{
if(st[v][j]==1)
{
st[u][j] = 1;
st[j][u] = -1;
}
if(st[u][j] == -1)
{
st[j][v] = 1;
st[v][j] = -1;
}
}
}
for(int i = 1; i <= n; i ++)
for(int j = 1; j <= n; j ++)
if(st[i][j] == 0)
for(int k = 1; k <= n; k ++)//通过k能够把i、j关系确定
{
if(st[i][k] == 1 && st[k][j] == 1)
st[i][j] = 1,st[j][i] = -1;
if(st[j][k] == 1 && st[k][i] == 1)
st[j][i] =1,st[i][j] = -1;
}
int ans = 0;
for(int i = 1; i <= n; i ++)
{
bool flag = true;
for(int j = 1; j <= n; j++)
if(st[i][j]==0&&i!=j)flag = false;
if(flag)ans++;
}
cout << ans;
return 0;
}
注意点
1.通过k能够把i、j关系确定