poj 1789 Truck History

来源：互联网发布：linux ansi 编辑：程序博客网时间：2024/06/09 14:04

prim求最小生成树。

 1 #include<stdio.h> 2 #include<string.h> 3 #define INF 100000000 4 #define MAXN 2100 5  6 int n, v[MAXN], low[MAXN], d[MAXN][MAXN]; 7 char s[MAXN][10]; 8  9 void prim()10 {11     int ans = 0;12     memset(v, 0, sizeof(v));13     v[0] = 1;14     for(int i = 1; i < n; i ++) low[i] = d[0][i];15     for(int i = 1; i < n; i ++)16     {17         int min = INF, p;18         for(int j = 0; j < n; j ++)19         if(!v[j] && min > low[j]) min = low[p=j];20         ans += min;21         v[p] = 1;22         for(int j = 0; j < n; j ++)23         if(!v[j] && low[j] > d[p][j]) low[j] = d[p][j];24     }25     printf("The highest possible quality is 1/%d.\n",ans);26 }27 void init()28 {29     while(scanf("%d",&n))30     {31         if(n == 0) break;32         for(int i = 0; i < n; i ++)33             scanf("%s",s[i]);34         for(int i = 0; i < n; i ++)35         {36             for(int j = i+1; j < n; j ++)37             {38                 int t = 0;39                 for(int k = 0; k < 7; k ++)40                     if(s[i][k] != s[j][k]) t ++;41                 d[i][j] = t;42                 d[j][i] = t;43             }44         }45         prim();46     }47 }48 int main()49 {50     init();51     return 0;52 }

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