HDU 1162 Eddy's picture （最小生成树 普里姆 ）

题目链接

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

分析：

题上要求的是用最短的线将坐标系上的n个点链接起来，这就是简单的最小生成树的问题，将任意两个点之间的距离计算出来，看作他们之间的路径长度，这样的话就能很好的理解为什么是最小生成树的问题了。

代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#define INF 1 << 30
double a[101] , b[101] , map[101][101] ;
double dis[101] ;
int used[101] ;
void prim(int n)
{
    int c = 0 ;
    int i = 0 , j = 0 ;
    double sum = 0 ;
    dis[1] = 0 ;
    for(i = 1 ; i <= n ; i++)
    {
        double min = INF ;
        c = 0 ;
        for(j = 1 ; j <= n ; j++)
        {
            if(!used[j] && dis[j] < min)
            {
                min = dis[j] ;
                c =j ;
            }
        }
        used[c] = 1 ;
        for(j  = 1 ; j <= n ; j++ )
        {
            if(!used[j] && dis[j] > map[c][j])
                dis[j] = map[c][j] ;
        }
    }

    for(i = 1 ; i <= n ; i++)
        sum += dis[i] ;
    printf("%.2lf\n",sum);
}
int main()
{
    int n = 0 ;
    while(~scanf("%d",&n))
    {
        memset(a , 0 , sizeof( a ) ) ;
        memset(b , 0 , sizeof( b ) ) ;
        int i = 0 , j = 0 ;
        for(i = 1 ; i <= n ; i++)
        {
            for(j = 1 ; j <= n ; j++)
                map[i][j] = INF ;
            dis[i] = INF ;
            used[i] = 0 ;
        }
        for(i = 1 ; i <= n ; i++)
        {
            scanf("%lf%lf" , &a[i] , &b[i] );///切记不能向一般的图论的问题，在输入的同时保存路径
        }
        double m = 0 , x = 0;
        for(i = 1 ; i <= n ; i++ )///应该在输入结束之后，将任意两点之间的路径计算出来
        {
            for(j = 1 ; j <= n ; j++)
            {
                x = (a[j]-a[i])*(a[j]-a[i])+(b[j]-b[i])*(b[j]-b[i]) ;
                m = sqrt( x ) ;
                map[i][j] = map[j][i] = m ;
            }
        }
        prim( n );
    }
    return 0 ;
}