poj 3660 Cow Contest
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Cow Contest
题目链接:http://poj.org/problem?id=3660
题目大意:求出能够确定排名的牛的个数。
有点类似于闭包传递,转化成图的连通性。最后统计各点的出入度等于n-1的便是符合要求的。
代码如下:
#include<cstdio>#include<iostream>#include<cstring>#define INF 100000000using namespace std;int map[105][105];int dis[105],vi[105];int ans[105];void Dj(int x,int n){for(int i=1;i<=n;i++){dis[i]=map[x][i];vi[i]=0;}vi[x]=1;for(int i=2;i<=n;i++){int p;for(int j=1;j<=n;j++){if(!vi[j] && dis[j]<INF){p=j;break;}}vi[p]=1;for(int j=1;j<=n;j++){if(!vi[j] && dis[j]>=map[p][j])dis[j]=map[p][j];}}}int main(){int n,m;int x,y;int sum;while(~scanf("%d%d",&n,&m)){sum=0;for(int i=0;i<=n;i++)ans[i]=0;for(int i=0;i<=n;i++)for(int j=0;j<=n;j++)map[i][j]=INF;for(int i=0;i<m;i++){cin>>x>>y;map[x][y]=1;}for(int i=1;i<=n;i++){Dj(i,n);for(int j=1;j<=n;j++){if(dis[j]<INF){ans[j]++;ans[i]++;}}}for(int i=1;i<=n;i++){if(ans[i]==n-1){sum++;}}printf("%d\n",sum);}return 0;}
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