题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1162

Eddy‘s picture

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6866    Accepted Submission(s): 3469

Problem Description

Eddy begins to like painting pictures recently ,he is sure of himself to become a painter.Every day Eddy draws pictures in his small room, and he usually puts out his newest pictures to let his friends appreciate. but the result it can be imagined, the friends are not interested in his picture.Eddy feels very puzzled,in order to change all friends ‘s view to his technical of painting pictures ,so Eddy creates a problem for the his friends of you. Problem descriptions as follows: Given you some coordinates pionts on a drawing paper, every point links with the ink with the straight line, causes all points finally to link in the same place. How many distants does your duty discover the shortest length which the ink draws?

Input

The first line contains 0 < n <= 100, the number of point. For each point, a line follows; each following line contains two real numbers indicating the (x,y) coordinates of the point.
Input contains multiple test cases. Process to the end of file.

Output

Your program prints a single real number to two decimal places: the minimum total length of ink lines that can connect all the points.

Sample Input

3

1.0 1.0

2.0 2.0

2.0 4.0

Sample Output

3.41

题目大意：用最短的线段，使所有的点连通。也就是最短路的一种应用，不多说，prim就可以解决了~~特别注意是输入多组数据的，否则会wa的！！！

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 using namespace std;
 5 double node[110],map[110][110],Min,n;
 6 const double INF=999999999;
 7
 8 double prim()
 9 {
10     int vis[110]= {0};
11     int tm=1,m;
12     double sum=0;
13     vis[tm]=1;
14     node[tm]=0;
15     for (int k=2; k<=n; k++)
16     {
17         Min=INF;
18         for (int i=1; i<=n; i++)
19             if (!vis[i])
20             {
21                 if (node[i]>map[tm][i])
22                     node[i]=map[tm][i];
23                 if (Min>node[i])
24                 {
25                     Min=node[i];
26                     m=i;
27                 }
28             }
29         vis[m]=1;
30         tm=m;
31         sum+=node[m];
32     }
33     //for (int i=1; i<=n; i++)
34     //sum+=node[i];
35     return sum;
36 }
37
38 int main ()
39 {
40     double a[110],b[110];
41     while (cin>>n)
42     {
43         for (int i=1; i<=n; i++)
44             cin>>a[i]>>b[i];
45         for (int i=1; i<=n; i++)
46         {
47             node[i]=INF;
48             for (int j=1; j<=n; j++)
49                 map[i][j]=INF;
50         }
51         for (int i=1; i<=n; i++)
52         {
53             for (int j=1; j<=n; j++)
54             {
55                 map[i][j]=map[j][i]=sqrt((a[i]-a[j])*(a[i]-a[j])+(b[i]-b[j])*(b[i]-b[j]));
56             }
57         }
58         printf ("%.2lf\n",prim());
59     }
60     return 0;
61 }

hdu 1162 Eddy's picture(最小生成树算法)