poj 1948 Triangular Pastures

Triangular Pastures

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 6475		Accepted: 2108

Description

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. 

Input

* 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. 

Output

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. 

Sample Input

5
1
1
3
3
4

Sample Output

692

Hint

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

Source

USACO 2002 February

题意：给n个棍，每个棍都有长度，求用这n个棍能摆成的最大三角形面积。

二维01背包 dp[i][j]表示两边分别为i和j的状态是否可达，这里i>=j，避免重复浪费空间，面积用海伦公式求，求的过程中要用double存，状态转移在代码中了

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
bool dp[805][805];
int a[41],sum;
int area(int i,int j,int k)
{
    double p=sum/2.0;
    return int(sqrt(p*(p-i)*(p-j)*(p-k))*100);
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF){
        int ans=-1;
        sum=0;
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        int m=(sum+1)/2;
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;
        for(int i=1;i<=n;i++)
            for(int j=m;j>=0;j--)
                for(int k=j;k>=0;k--)
                    if(dp[j][k])dp[j+a[i]][k]=dp[j][k+a[i]]=1;

        for(int j=m;j>=0;j--)
            for(int k=j;k>=0;k--)
                if(dp[j][k])ans=max(ans,area(j,k,sum-j-k));

        printf("%d\n",ans);
    }
    return 0;
}