Eddy's picture 杭电oj 1162

最新推荐文章于 2020-06-15 08:06:49 发布

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#Eddys picture 杭电oj 1

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本篇介绍了一个有趣的绘画问题：如何用最短的总距离将平面上的若干点通过直线相连。通过普利姆算法求解最小生成树，实现对绘画问题的有效解决。

ddy's picture

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

Total Submission(s): 5844 Accepted Submission(s): 2949

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 Outpu

3.41

/*普利姆算法求最小生成树*/
#include<iostream>
#include<iomanip>
#include<cmath>
using namespace std;
const int MAX = 101;
const double INF = 1000000.0;
struct Node
{
double x;
double y;
}node[MAX]; //顶点的定义表
double closedge[MAX]; //辅助变量
int main()
{
int n;
while(cin >> n)
{
int i,j;
for(i = 0;i < n;i++)
cin >> node[i].x >> node[i].y;
for(j = 0;j < n;j++)
closedge[j] = INF;
int v = 0;
double sum = 0;
closedge[v] = 0;
for(i = 0;i < n;i++)
{
for(j = 0;j < n;j++)
{
double dis = sqrt((node[v].x-node[j].x)*(node[v].x-node[j].x)+(node[v].y-node[j].y)*(node[v].y-node[j].y));
if(j != v&&dis < closedge[j])
closedge[j] = dis;
}
sum += closedge[v];
closedge[v] = 0;
double max = INF;
for(j = 0;j < n;j++)
if(closedge[j]>0&&closedge[j]<max)
{
max = closedge[j];
v = j;
}
}
cout << setiosflags(ios::fixed) << setprecision(2) << sum << endl;
}
return 0;
}