poj1948 Triangular Pastures（背包）

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

分析：
我们可以通过两条边得出第三条边的长度
那么我们只用看看两条边怎么分配就好了
这就有点像一个背包，但是只判断可行性
还是枚举每一条line以及组成的两条边的长度
f[j][k] 一条边是j，一条边是k，那么剩下的一条边就是sum-k-j
f[j][k]=f[j][k]||f[j-a[i]][k]
f[j][k]=f[j][k]||f[j][k-a[i]]

三边求出来之后
我们要怎么计算三角形面积呢：

这里写代码片
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>

using namespace std;

int ans=-1;
int n,sum=0;
int a[100];
bool f[2000][2000];

void js(int x,int y)
{
    int z=sum-x-y;
    if (x+y>z&&y+z>x&&x+z>y)
    {
        double S=(double)(x+y+z)*(x+y-z)*(x+z-y)*(y+z-x);
        S=sqrt(S); S=S/4;
        S=S*100;
        ans=max(ans,(int)S);
    }
}

int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++) scanf("%d",&a[i]),sum+=a[i];
    int i,j,k;
    f[0][0]=1;
    for (i=1;i<=n;i++)
        for (j=sum;j>=0;j--)
            for (k=sum-j;k>=0;k--)
            {
                if (j>=a[i]) f[j][k]=f[j][k]||f[j-a[i]][k];
                if (k>=a[i]) f[j][k]=f[j][k]||f[j][k-a[i]];
                if (f[j][k]) js(j,k);
            }
    printf("%d",ans);
    return 0;
}