ZOJ 1203 Swordfish

 1 #include <cstdio>
 2 #include <cmath>
 3 
 4 double dis[105][105];
 5 double p[105][2]; //输入坐标点
 6 double d[105];  //一点到另一点的距离
 7 double vis[105];  //记得该点是否已被访问
 8 double len;
 9 int n;
10 
11 //初始化dis[105][105]
12 void init()
13 {
14     int i,j;
15     for(i=1; i<=n; i++)
16     {
17         for(j=1; j<=n; j++)
18         {
19             dis[i][j]=dis[j][i]=100000000;
20         }
21         vis[i] = 0;
22     }
23     len=0.0;
24 }
25 
26 //计算dis[105][105]
27 void makedis()
28 {
29     int i,j;
30 
31     for(i = 1; i <= n; i++)
32         for(j = i; j <= n; j++)
33         {
34             dis[i][j] = dis[j][i] = sqrt((p[i][0]-p[j][0])*(p[i][0]-p[j][0])+(p[i][1]-p[j][1])*(p[i][1]-p[j][1]));
35 
36         }
37 }
38 
39 
40 //用prim算法求最短距离
41 void prim()
42 {
43     int i,j,k;
44     double min;
45     for(i = 1; i <= n; i++)
46         d[i] = dis[1][i];
47     vis[1] = 1;
48 
49     for( i = 1; i < n; i++)
50     {
51         min = 0xfffffff;              // 最大值
52         for ( j = 1; j <= n; j++)
53         {
54             if ( d[j] < min && !vis[j])
55             {
56                 min = d[j];
57                 k = j;
58             }
59         }
60         vis[k] = 1;
61         len += d[k];
62         for(j = 1; j <= n; j++)
63         {
64             if(!vis[j] && d[j] > dis[k][j])
65                 d[j] = dis[k][j];
66         }
67 
68     }
69 }
70 
71 //主函数
72 int main()
73 {
74     int step;
75     int i;
76     step=0;
77     while(scanf("%d",&n) && n)     //以0结束
78     {
79         step++;              //记录case
80 
81         if(step!=1)
82             printf("
");
83         init();
84         for(i = 1; i <= n; i++)
85         {
86             scanf("%lf%lf",&p[i][0],&p[i][1]);
87         }
88         makedis();
89         prim();
90         printf("Case #%d:
",step);
91         printf("The minimal distance is: %.2lf
",len);
92     }
93     return 0;
94 }