题目链接
题意
n头牛,给出m个a打败b的结果,求有多少头牛的排名可以被确定。
思路
能确定排名的牛,也就是能确定击败它的牛的数量加上能确定被它击败的数量等于n-1,就可以确定排名。
只需要判断能否到达,直接用floyd算法:
mp[i][j]|=(mp[i][k]&mp[k][j]);
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string.h>
using namespace std;
const int maxn = 300;
typedef long long ll;
int n, m, mp[maxn][maxn];
int main()
{
cin >> n >> m;
memset(mp, 0, sizeof(mp));
for (int i = 1; i <= m; i++) {
int x, y;
cin >> x >> y;
mp[x][y] = 1;
}
for (int k = 1; k <= n; k++) {
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
mp[i][j] |= (mp[i][k]&mp[k][j]);
}
int ans = 0;
for (int i = 1; i <= n; i++) {
int a = 0, b = 0;
for (int j = 1; j <= n; j++) a += mp[i][j], b += mp[j][i];
if(a + b == n - 1) ans++;
}
printf("%d\n", ans);
return 0;
}