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;
}
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