poj 1948 Triangular Pastures(2RD 01 가방)

Triangular Pastures
Time Limit: 1000MS
 
Memory Limit: 30000K
Total Submissions: 4333
 
Accepted: 1295
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
제목:http://poj.org/problem?id=1948
분석: 이 문제는 2차원 01 가방 문제라고 할 수 있습니다. 상태 이동이 뚜렷합니다. f[i][j]는 첫 번째 변의 길이가 i이고 두 번째 변의 길이가 j라고 가정하면 기본 i>=j, f[i][j]=f[j-l[k]][j]|f[j][j=l[k](1<=k<=n)입니다. 이 문제에 주의하는 점은 면적은 p=(a+b+c)/2.0입니다.이렇게 5++번~
코드:

#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
bool f[888][888];
int l[44];
int area(int x,int y,int z)
{
    double p=(x+y+z)/2.0;
    return int(sqrt(p*(p-x)*(p-y)*(p-z))*100);
}
int main()
{
    int i,j,k,n,m,sum,ans;
    while(scanf("%d",&n)!=-1)
    {
        for(sum=i=0;i<n;++i)scanf("%d",&l[i]),sum+=l[i];
        m=sum/2-((sum&1)==0);
        for(i=0;i<=m;++i)
            for(j=0;j<=i;++j)f[i][j]=0;
        f[0][0]=1;
        for(k=0;k<n;++k)
            for(i=m;i>=0;--i)
                for(j=i;j>=0;--j)
                    if(f[i][j])f[i+l[k]][j]=f[i][j+l[k]]=1;
        ans=-1;
        for(i=1;i<=m;++i)
            for(j=1;j<=i;++j)
                if(f[i][j])ans=max(ans,area(i,j,sum-i-j));
        printf("%d
",ans);
    }
    return 0;
}