HDU5135 Little Zu Chongzhi's Triangles(计算几何，枚举)

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题目：

Little Zu Chongzhi's Triangles

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 1901 Accepted Submission(s): 1083

Problem Description

Zu Chongzhi (429–500) was a prominent Chinese mathematician and astronomer during the Liu Song and Southern Qi Dynasties. Zu calculated the value ofπ to the precision of six decimal places and for a thousand years thereafter no subsequent mathematician computed a value this precise. Zu calculated one year as 365.24281481 days, which is very close to 365.24219878 days as we know today. He also worked on deducing the formula for the volume of a sphere.

It is said in some legend story books that when Zu was a little boy, he liked mathematical games. One day, his father gave him some wood sticks as toys. Zu Chongzhi found a interesting problem using them. He wanted to make some triangles by those sticks, and he wanted the total area of all triangles he made to be as large as possible. The rules were :

1) A triangle could only consist of 3 sticks.
2) A triangle's vertexes must be end points of sticks. A triangle's vertex couldn't be in the middle of a stick.
3) Zu didn't have to use all sticks.

Unfortunately, Zu didn't solve that problem because it was an algorithm problem rather than a mathematical problem. You can't solve that problem without a computer if there are too many sticks. So please bring your computer and go back to Zu's time to help him so that maybe you can change the history.

Input

There are no more than 10 test cases. For each case:

The first line is an integer N(3 <= N<= 12), indicating the number of sticks Zu Chongzhi had got. The second line contains N integers, meaning the length of N sticks. The length of a stick is no more than 100. The input ends with N = 0.

Output

For each test case, output the maximum total area of triangles Zu could make. Round the result to 2 digits after decimal point. If Zu couldn't make any triangle, print 0.00 .

Sample Input

  3
1 1 20
7
3 4 5 3 4 5 90
0

Sample Output

  0.00
13.64

Source

2014ACM/ICPC亚洲区广州站-重现赛（感谢华工和北大）

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Statistic | Submit | Discuss | Note 思路：
有n个数，接下来有n个小木棍，要把它尽量围成多个三角形，并且使，围成的三角形的面积最大。

需要满足三个条件：

①三角形只能有三根棒组成

②三角形的顶点必须是棒的终点

③木棍不能用完

先按照贪心的思路给边排序，然后在枚举

代码：

#include <cstdio>
#include <cstring>
#include <cctype>
#include <string>
#include <set>
#include <iostream>
#include <stack>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
#define mod 10000007
#define debug() puts("what the fuck!!!")
#define N 100
#define M 1000000
#define ll long long
using namespace std;
struct node
{double l;int vis;
} a[N];
int cmp(node a,node b)
{return a.l>b.l;
}
double mj(double a,double b,double c)
{double p=(a+b+c)/2;return sqrt(p*(p-a)*(p-b)*(p-c));
}
int jc(double a,double b,double c)
{if(a+b>c&&a+c>b&&b+c>a)return 1;elsereturn 0;
}
int main()
{int n;while(scanf("%d",&n)&&n){for(int i=1; i<=n; i++){scanf("%lf",&a[i].l);a[i].vis=0;}sort(a+1,a+n+1,cmp);double num=0.0;int flag;for(int i=1; i<=n; i++){flag=0;if(a[i].vis==0&&flag==0){for(int j=i+1; j<=n; j++){if(a[j].vis==0&&flag==0){for(int k=j+1; k<=n; k++){if(a[k].vis==0&&flag==0){if(jc(a[i].l,a[j].l,a[k].l)){//debug();num+=mj(a[i].l,a[j].l,a[k].l);a[i].vis=1;a[j].vis=1;a[k].vis=1;flag=1;break;}}}}}}}printf("%.2lf\n",num);}return 0;
}

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