HD 1162 Eddy's picture

Eddy's picture

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 11578    Accepted Submission(s): 5839

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

解題思路：最小生成樹，類似與prim()算法

代碼如下：

# include<stdio.h>
# include<math.h>
# include<string.h>
struct node{
    double x,y;
}a[105],b[105];//b[]保存的是在最小生成樹中的點，a[]保存的不在最小生成樹的點
int v[105];
double len(node a,node b)  //計算兩點的距離
{
    return sqrt((a.x-b.x)*(a.x-b.x)*1.0+(a.y-b.y)*(a.y-b.y));
}
int main(){
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
        scanf("%lf%lf",&a[i].x,&a[i].y);    
        }
        double sum=0;
        memset(v,0,sizeof(v));
        v[1]=1;
        int k=1;
        b[k++]=a[1]; 
        while(k<=n)
        {  double min=0x3f3f;
        int kk;
            for(int i=1;i<=n;i++)
            {
                if(!v[i])  //如果改點不在最小生成樹中
                {
                    for(int j=1;j<k;j++)  //找到 到 最小生成樹中點的最短距離的點
                    {
                        if(len(a[i],b[j])-min<0.000001)
                        {
                        kk=i;     //kk就是 到 最小生成樹中點的最短距離的點
                        min=len(a[i],b[j]);    
                        
                        }                        
                    }                                        
                }                                
            }
            v[kk]=1;
            sum+=min;
            b[k++]=a[kk];  //將kk這個點加入最小生成樹中
        }
        printf("%.2f\n",sum);
    }
    
    return 0;
}