Truck History(prim)

http://poj.org/problem?id=1789

读不懂题再简单也不会做，英语是硬伤到哪都是真理，sad++。

此题就是一个最小生成树，两点之间的权值是毎两串之间的不同字母数。

 #include<stdio.h>
 #include<string.h>
 const int N=;
 const int INF=<<;
 int map[N][N],vis[N],dis[N];
 int n,sum;
 void prim()
 {
     int pos;
     for (int i = ; i <= n; i ++)
     {
         dis[i] = map[][i];
     }
     vis[] = ;
     for (int i = ; i <= n-; i ++)
     {
         int min = INF;
         for (int j = ; j <= n; j ++)
         {
             if (!vis[j] && dis[j] < min)
             {
                 min = dis[j];
                 pos = j;
             }
         }
         sum += min;
         vis[pos] = ;
         for (int j = ; j <= n; j ++)
         {
             if (!vis[j] && dis[j] > map[pos][j])
                 dis[j] = map[pos][j];
         }

     }
 }
 void init()
 {
     sum = ;
     for (int i = ; i <= n; i ++)
     {
         for (int j = ; j <= n; j ++)
         {
             map[i][j]  = INF;
         }
         map[i][i] = ;
         vis[i] = ;
     }

 }
 int main()
 {
     char str[N][];
     int cnt;
     while(~scanf("%d",&n)&&n)
     {

         init();
         for (int i = ; i <= n; i ++)
         {
             scanf("%s",str[i]+);
         }
         for (int i = ; i <= n; i ++)
         {
             for (int j = i+; j <= n; j ++)
             {
                 cnt = ;
                 for (int k = ; k <= ; k ++)
                 {
                     if (str[i][k]!=str[j][k])
                         cnt++;
                 }
                 map[i][j] = cnt;
                 map[j][i] = cnt;
             }
         }
         prim();
         printf("The highest possible quality is 1/%d.\n",sum);
     }
     return ;

 }