Kuskal/Prim POJ 1789 Truck History

题目传送门

 1 /*
 2     题意：给出n个长度为7的字符串，一个字符串到另一个的距离为不同的字符数，问所有连通的最小代价是多少
 3     Kuskal/Prim: 先用并查集做，简单好写，然而效率并不高，稠密图应该用Prim。这是最小生成数的裸题，然而题目有点坑爹:(
 4 */
 5 #include <cstdio>
 6 #include <cstring>
 7 #include <string>
 8 #include <algorithm>
 9 #include <iostream>
10 #include <cmath>
11 using namespace std;
12
13 const int MAXN = 2e3 + 10;
14 const int INF = 0x3f3f3f3f;
15 struct UF
16 {
17     int rt[MAXN];
18
19     void init(void)    {memset (rt, -1, sizeof (rt));}
20
21     int Find(int x)    {return (rt[x] == -1) ? x : rt[x] = Find (rt[x]);}
22
23     void Union(int x, int y)
24     {
25         x = Find (x);    y = Find (y);
26         if (x < y)    rt[x] = y;
27         else    rt[y] = x;
28     }
29
30     bool same(int x, int y)    {return (Find (x) == Find (y));}
31 }uf;
32 char s[MAXN][10];
33 struct Node
34 {
35     int u, v, w;
36 }node[MAXN*MAXN/2];
37 int n, tot;
38
39 bool cmp(Node x, Node y)    {return x.w < y.w;}
40
41 void get_w(int x)
42 {
43     for (int i=1; i<x; ++i)
44     {
45         int res = 0;
46         for (int j=0; j<7; ++j)
47         {
48             if (s[i][j] != s[x][j])    res++;
49         }
50         node[++tot].u = i;    node[tot].v = x;    node[tot].w = res;
51     }
52 }
53
54 int main(void)        //POJ 1789 Truck History
55 {
56     // freopen ("POJ_1789.in", "r", stdin);
57
58     while (scanf ("%d", &n) == 1)
59     {
60         if (n == 0)    break;
61
62         tot = 0;
63         for (int i=1; i<=n; ++i)
64         {
65             scanf ("%s", s[i]);
66             get_w (i);
67         }
68         sort (node+1, node+1+tot, cmp);
69
70         int ans = 0;    uf.init ();
71         for (int i=1; i<=tot; ++i)
72         {
73             int u = node[i].u;    int v = node[i].v;    int w = node[i].w;
74             if (!uf.same (u, v))    {uf.Union (u, v);    ans += w;}
75         }
76
77         printf ("The highest possible quality is 1/%d.\n", ans);
78     }
79
80     return 0;
81 }