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最新推荐文章于 2020-06-13 09:39:41 发布

转载最新推荐文章于 2020-06-13 09:39:41 发布 · 111 阅读

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原文链接：http://www.cnblogs.com/playboy307/p/5273838.html

文章标签：

#java

本文探讨了一个在给定坐标点上绘制最短连线的问题，通过使用并查集算法来解决如何将所有点连接在一起，形成最小总长度的连线。详细介绍了输入输出规范及示例代码。

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

Total Submission(s) : 8 Accepted Submission(s) : 3

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

Sample Output

3.41

Author

eddy

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
int pre[101];
void itin()
{
    for(int i=0;i<101;i++)
    pre[i]=i;
}
struct point
{
    float x,y;
}p[100];
struct dis
{
    int u,v;
    float val;
}d[100010];
int cmp(dis n1,dis n2)
{
    if(n1.val<n2.val)
    return 1;
    return 0;
}
int find(int x)
{
    while(x!=pre[x])
    x=pre[x];
    return x;
}
bool join(int x,int y)
{
    int fx=find(x);
    int fy=find(y);
    if(fx!=fy)
    {
        pre[fx]=fy;
        return true;
    }
    return false;
}
int main()
{
    int t;
    while(scanf("%d",&t)!=EOF)
    {
        itin();
        int i,j;
        memset(p,0,sizeof(p));
        float sum=0;
        for(i=0;i<t;i++)
        scanf("%f%f",&p[i].x,&p[i].y);
        int cnt=0;
        for(i=0;i<t;i++)
        for(j=i+1;j<t;j++)
        {
            float dist=sqrt((p[i].x-p[j].x)*(p[i].x-p[j].x)+(p[i].y-p[j].y)*(p[i].y-p[j].y));
            d[cnt].u=i;
            d[cnt].v=j;
            d[cnt++].val=dist;
        }
        sort(d,d+cnt,cmp);
        /*for(i=0;i<cnt;i++)
        printf("%f ",d[i].val);
        printf("\n");*/
        for(i=0;i<cnt;i++)
        {
            if(join(d[i].u,d[i].v))
            {
            sum+=d[i].val;
            //printf("%f\n",sum);
            }
            
        }
        printf("%.2f\n",sum);
    }
    return 0;
}

转载于:https://www.cnblogs.com/playboy307/p/5273838.html