poj1948 二维01背包

http://poj.org/problem?id=1948

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

/**
题目大意；给定一些长度不等的木棍，求有这些木棍能拼出的三角形的最大面积是多少？
解题思路：二维01背包问题。dp[i][j]表示有两条边为i和j的三角形，用背包列出所有可行的状态，然后比较面积取最大。
由于三角形的边长小于周长的一半，我们从sum/2开始枚举，这样就将时间复杂度降低到1s内可求的范围。
*/
#include 
#include 
#include 
#include 
#include 
using namespace std;

int a[45],dp[900][900];
int n;

int main()
{
    while(~scanf("%d",&n))
    {
        int sum=0;
        for(int i=0; i>1;
        for(int i=0; i=0; j--)
            {
                for(int k=j; k>=0; k--)
                {
                    if(j>=a[i]&&dp[j-a[i]][k]||dp[j][k-a[i]]&&k>=a[i])
                    {
                        dp[j][k]=1;
                    }
                }
            }
        }
        int ans=-1;//这里初始化为0就是错，为什么？？
        for(int i = half; i >= 1; i--)
        {
            for(int j = i; j >= 1; j--)
            {
                if(dp[i][j])
                {
                    int k=sum-i-j;
                    if(i+j>k&&i+k>j&&j+k>i)
                    {
                        double p = (i + j + k) * 1.0 / 2;
                        int temp = (int)(sqrt(p * (p - i) * (p - j) * (p - k)) * 100);
                        if (temp > ans)
                            ans = temp;
                    }
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}