题目：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3716

来源：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=26644#problem/A

Ribbon Gymnastics

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Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge

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Robert is a gymnastics coach. Unfortunately, he got four gymnastics beginners to attend the coming competition. Since the competition has a nice award, Robert will make all his effort
to win the competition.

One performance requires that each of the four player should take a ribbon and rotate herself, and assume the ribbons will then straighten out. Since Robert's four players are beginners,
Robert cannot control the actual rotating speeds of four players during the competition. And when two ribbons touch, they may wind, which will cause players felling down. For safety, Robert should avoid this. Four distinct points will be given as x_i y_i before
the competition begins. Players should stand at those points when rotating.

The longer the ribbons are, the more bonus Robert will gain. After knowing the four points Robert can choose ribbons and points for each player. The length of each ribbon should be positive.
So help Robert to find the maximal total lengths of the four ribbons.

Input

There will be multiple test cases.Please process to the end of input.

Each test case contains four lines.

Each line contains two integers x_i (-10^8≤x_i≤10^8) and y_i (-10^8≤y_i≤10^8).

Output

Output the total maximal sum of the lengths of the four ribbons. Answer having an absolute or relative error less than 1e-6 will be accepted.

Sample Input

0 0

0 3

4 0

4 3

Sample Output

6.00000000

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Author: YANG, Jueji

Contest: ZOJ Monthly, June 2013

题意：给你四个点，四个人在四个点上舞动丝带，各条丝带不能相遇，求四人最大的丝带总长度。

思路：开始比赛时想的是以这四个点画圆求四个圆尽量相切，求最大的半径和。

　　　先找出两两点中的最短的一条线，由此必然确定了两个相切的圆，然后再上下移动，慢慢的确定另外的圆。

一直没想到怎么确定，后来就放弃了。。。

刚刚搜了下题解，应该这个是正解，也是圆相切的http://blog.csdn.net/u010638776/article/details/9218995

刚刚问了下KB神，他给了个传说中世界冠军 ＹＹ的思路

迅速秒过啊，求知道的大神给个证明Ｏｒｚ

A

Accepted

180 KB

0 ms

C++ (g++ 4.4.5)

729 B

2013-07-20 17:27:52

#include<stdio.h>

#include<math.h>

#include<algorithm>

using namespace std;

struct Point{

    double x;

    double y;

}p[5];

double dist(Point a, Point b)

{

    return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));

}

int main()

{

    while(scanf("%lf%lf", &p[0].x, &p[0].y) != EOF)

    {

        for(int i = 1; i < 4; i++)

        {

            scanf("%lf%lf", &p[i].x, &p[i].y);

        }

        double d1 = dist(p[0], p[1]);

        double d2 = dist(p[1], p[2]);

        double d3 = dist(p[2], p[3]);

        double d4 = dist(p[0], p[3]);

        double d5 = dist(p[0], p[2]);

        double d6 = dist(p[1], p[3]);

        double ans = min(d1+d3, min(d4+d2, d5+d6));

        printf("%.8lf\n", ans);

    }

    return 0;

}