Six Degrees of Cowvin Bacon.(POJ-2139)
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一道简单的图论题,不过穿上了很好的外衣,实质就是一个任意两点间最短路问题,比较适合用Floyd算法
#include<cstdio>#include<iostream>using namespace std;const int INF = 100000;int n,m,t,d[305][305],a[305];int main() { scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) if(i==j) d[i][j] = 0; else d[i][j] = d[j][i] = INF;//预处理,将任意两点的距离设为INF,但是d[i][i] = 0; for(int i=0;i<m;i++) { scanf("%d",&t); for(int j=0;j<t;j++) scanf("%d",&a[j]); for(int j=0;j<t;j++) for(int k=0;k<t;k++) { if(j!=k) { d[a[j]][a[k]] = 1; d[a[k]][a[j]] = 1; } } } for(int k=1;k<=n;k++) for(int i=1;i<=n;i++) <span style="white-space:pre"></span> for(int j=1;j<=n;j++) d[i][j] = min(d[i][j],d[i][k] + d[k][j]);//Floyd int sum,maxn = 1000000; for(int i=1;i<=n;i++) { sum = 0; for(int j=1;j<=n;j++) if(i!=j) sum += d[i][j]; if(sum<maxn) maxn = sum; } printf("%d\n",(maxn*100)/(n-1)); return 0;}
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