Program4_B

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本文介绍了一道关于计算几何的问题，即如何通过连线使得n个二维平面上的点彼此连通且连线总长度最短。文章提供了使用最小生成树算法解决此问题的详细步骤和示例代码。

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我现在做的是第四专题编号为1002的试题，主要内容是：

Problem B

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 38 Accepted Submission(s) : 8

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

简单题意：
给出n个点的坐标（二维x， y），可以使用直线（无向的）将任意两个点连接起来，求将所有点连接起来形成一个整体（使任何两点之间可达），线的最小距离。

解题思路：
最小生成树，用了两点间的距离公式，普利姆算法解决问题

编写代码：

#include<stdio.h>
#include<cstring>
#include<math.h>
double map[105][105];
int v[105];
double lowcost[105];
int n;
void init()
{
 int i,j;
 memset(v,0,sizeof(v));
 for(i=0; i<n; i++)
 for(j=0; j<n; j++)
 map[i][j]=1.7976931348623158e+308;
}
double prim(int u)
{
 int i,j;
 double min,mincost=0;
 v[u]=1;
 for(i=0; i<n; i++)
 lowcost[i]=map[u][i];
 for(i=1; i<n; i++)
 {
 min=1.7976931348623158e+308;
 for(j=0; j<n; j++)
 if(lowcost[j]<min && !v[j])
 min=lowcost[j],u=j;
 v[u]=1;
 mincost+=min;
 for(j=0; j<n; j++)
 if(lowcost[j]>map[u][j] && !v[j])
 lowcost[j]=map[u][j];
 }
 return mincost;
}
int main(){
 int i,j;
 double a[105],b[105];
 while(~scanf("%d",&n) && n) {
 init();
 for(i=0; i<n; i++)
 scanf("%lf%lf",a+i,b+i);
 for(i=0; i<n; i++)
 for(j=i; j<n; j++)
 {
 map[j][i]=map[i][j]=sqrt((a[i]-a[j])*(a[i]-a[j])+(b[i]-b[j])*(b[i]-b[j]));
 }
 printf("%.2f\n",prim(0));
 }
 return 0;
}