Triangular Pastures （二维01背包）

描述
Like everyone, cows enjoy variety. Their current fancy is new shapes for pastures. The old rectangular shapes are out of favor; new geometries are the favorite.

I. M. Hei, the lead cow pasture architect, is in charge of creating a triangular pasture surrounded by nice white fence rails. She is supplied with N (3 <= N <= 40) fence segments (each of integer length Li (1 <= Li <= 40) and must arrange them into a triangular pasture with the largest grazing area. Ms. Hei must use all the rails to create three sides of non-zero length.

Help Ms. Hei convince the rest of the herd that plenty of grazing land will be available. Calculate the largest area that may be enclosed with a supplied set of fence segments.

输入

* Line 1: A single integer N

* Lines 2..N+1: N lines, each with a single integer representing one fence segment's length. The lengths are not necessarily unique.

输出

A single line with the integer that is the truncated integer representation of the largest possible enclosed area multiplied by 100. Output -1 if no triangle of positive area may be constructed.

样例输入

5
1
1
3
3
4

样例输出

692

提示

which is 100x the area of an equilateral triangle with side length 4

题目大意：

给定几条边，用上全部的边组成三角形，输出最大的三角形面积，若不存在，输出-1.
大体思路：

枚举所有可能出现的边的情况，在判断是否能成为三角形，然后进行处理

#include <bits/stdc++.h>
using namespace std;
const int N=1605;///40*40
int dp[N][N]={0};///dp[i][j]代表能否构成i,j两边
int a[50];
int area(int a,int b,int c)
{
    double p=(a+b+c)/2.0;
    return (int)100*sqrt(p*(p-a)*(p-b)*(p-c));
}
int main()
{
    int n,sum=0;
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>a[i],sum+=a[i];
    dp[0][0]=1;
    for(int i=0;i<n;i++)
        for(int j=sum;j>=0;j--)
            for(int k=j;k>=0;k--)
                if(j>=a[i]&&dp[j-a[i]][k]||k>=a[i]&&dp[j][k-a[i]])
                    dp[j][k]=1;
    int maxs=-1;
    for(int i=sum;i>0;i--)
        for(int j=i;j>0;j--)
        {
            if(dp[i][j])
            {
                int k=sum-i-j;
                if(i<j+k||i+j>k)///判断三边是否组成三角形
                    maxs=max(maxs,area(i,j,k));
            }
        }
    if(maxs==-1) cout<<"-1"<<'\n';
    else cout<<maxs<<'\n';
    return 0;
}