poj1789Truck History(简单最小生成树）

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

模板题 题意是一个字符串跟一个字符不同的字符串有多少 这个数量就相当于权值 连起来最小

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 1 #include <iostream>

 2 #include<algorithm>

 3 #include<string.h>

 4 #include<cstdio>

 5 #define INF 0x3f3f3f3f

 6 using namespace std;

 7 char c[2001][10];

 8 int w[2001][2001],vis[2001],low[2001],s;

 9 void prime(int n)

10 {

11     int i,j,k,m;

12     memset(vis,0,sizeof(vis));

13     vis[1] = 1;

14     for(i =2 ; i <= n ; i++)

15     low[i] = w[1][i];

16     for(i = 1; i <= n ; i++)

17     {

18         m = INF;

19         for(j = 1; j <= n ; j++)

20         if(!vis[j]&&low[j]<m)

21         m = low[k=j];

22         if(m==INF)

23         break;

24         s+=m;

25         vis[k] = 1;

26         for(j = 1; j <= n ; j++)

27         if(!vis[j]&&low[j]>w[k][j])

28         low[j] = w[k][j];

29     }

30 }

31 int main()

32 {

33     int i,j,k,n,m,y,x;

34     while(scanf("%d%*c",&n)&&n)

35     {

36         memset(w,INF,sizeof(w));

37         s = 0;

38         for(i = 1; i <= n ; i++)

39         {

40             gets(c[i]);

41         }

42         for(i = 1; i <= n ; i++)

43         {

44             for(j = i+1; j <= n ; j++)

45             {

46                 y = 0;

47                 for(x = 0 ; x < 7 ; x++)

48                 if(c[j][x]!=c[i][x])

49                 y++;

50                 if(w[i][j]>y)

51                 {

52                     w[i][j] = y;

53                     w[j][i] = y;

54                 }

55             }

56         }

57         prime(n);

58         printf("The highest possible quality is 1/%d.\n",s);

59     }

60     return 0;

61 }