http://acm.nyist.net/JudgeOnline/problem.php?pid=211&&有向图传递闭包问题
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这一题是有向图的传递闭包问题,,,而这提的要求就是输出有多少牛可以确定名次的。算法思想:求出每个牛的出度和入度之和看是否等于n-1(n为牛的个数),我这里是用floyd算法写的,,,
#include<iostream>#include<algorithm>#include<string.h>#include<cstdio>#define N 105#define FOR(i,s,t) for(int i=(s);i<=(t);++i)using namespace std;bool map[N][N];int n,m;void floyd(){ FOR(k,1,n) FOR(i,1,n) FOR(j,1,n) if((map[i][k]&&map[k][j])||map[i][j]) map[i][j]=true; } int main() { while(~scanf("%d%d",&n,&m)!=EOF&&n&&m) { FOR(i,1,n)初始化工作 FOR(j,1,n) if(i==j) map[i][j]=true; else map[i][j]= false; FOR(i,1,m) { int a,b; scanf("%d%d",&a,&b); map[a][b]=true; } floyd(); int ans=0; FOR(i,1,n) { int sum=0; FOR(j,1,n) if(i==j) continue; else{ if(map[i][j]||map[j][i]) sum++; if(sum==n-1) ans++; } } printf("%d\n",ans); } return 0; }
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