POJ 1789 Truck History

两个字符串相同位置上不同的字符数代表两个节点间的距离，求把所有节点对都包括的路径的最小值的倒数

其实就是最小生成树

代码：

#include<iostream>using namespace std;#define inf 0x3f3f3f3f;char t[2001][10];int g[2001][2001],dis[2001],n;int min(int a,int b){return a<b?a:b;}void getDis(){int i,j,k;for(i=1;i<=n;i++)for(j=1;j<=n;j++){if(i==j)continue;for(k=0;k<7;k++){if(t[i][k]!=t[j][k])g[i][j]++;}}}int prim(){int i,j,now,min_w,min_node,ans;for(i=1;i<=n;i++)dis[i]=inf;now=1;ans=0;for(i=1;i<n;i++){dis[now]=-1;min_w=inf;for(j=1;j<=n;j++){if(now!=j&&dis[j]>=0){dis[j]=min(dis[j],g[now][j]);if(dis[j]<min_w){min_w=dis[j];min_node=j;}}}ans+=min_w;now=min_node;}return ans;}int main(){int i,j,q;while(scanf("%d",&n)!=EOF&&n){memset(g,0,sizeof(g));for(i=1;i<=n;i++)scanf("%s",&t[i]);getDis();q=prim();printf("The highest possible quality is 1/%d./n",q);}return 0;}