POJ 1789 Truck History【最小生成树模板题Kruscal】

题目链接：http://poj.org/problem?id=1789

大意：

不同字符串相同位置上不同字符的数目和是它们之间的差距。求衍生出全部字符串的最小差距。

 1 #include<stdio.h>
 2 #include<math.h>
 3 #include<algorithm>
 4 using namespace std;
 5 const int MAXN = 2100;
 6 
 7 int cnt;
 8 int pre[MAXN];
 9 char s[MAXN][10];
10 
11 struct Edge
12 {
13     int from, to;
14     int val;
15 }edge[MAXN * (MAXN - 1) / 2];
16 
17 bool cmp(Edge a, Edge b)
18 {
19     return a.val < b.val;
20 }
21 
22 int find(int x)
23 {
24     if(pre[x] == x)
25         return x;
26     else
27     {
28         int root = find(pre[x]);
29         pre[x] = root;
30         return pre[x];
31     }
32 }
33 
34 int main()
35 {
36     int n;
37     while(scanf("%d", &n) != EOF)
38     {
39         if(n == 0)
40             break;
41         getchar();
42         cnt = 0;
43         for(int i = 1; i <= n; i ++)
44             pre[i] = i;
45         for(int i = 1; i <= n; i ++)
46             scanf("%s", s[i] + 1);
47         for(int i = 1; i < n; i ++)
48         {
49             for(int j = i + 1; j <= n; j ++)
50             {
51                 int sum = 0;
52                 for(int k = 1; k <= 7; k ++)
53                     if(s[i][k] != s[j][k])
54                         sum ++;
55                 edge[++ cnt].from = i;
56                 edge[cnt].to = j;
57                 edge[cnt].val = sum;
58             }
59         }
60         sort(edge + 1, edge + 1 + cnt, cmp);
61         int ans = 0;
62         for(int i = 1; i <= cnt; i ++)
63         {
64             int xx = find(edge[i].from), yy = find(edge[i].to);
65             if(xx != yy)
66             {
67                 pre[yy] = xx;
68                 ans += edge[i].val;
69             }
70         }
71         printf("The highest possible quality is 1/%d.\n", ans);
72     }    
73     return 0;
74 }
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posted @ 2019-10-10 19:22  缘未到  阅读(95)  评论(0)  编辑  收藏  举报