Triangular Pastures

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]

题意：要构成三角形的牧场，当构成一个时，在接下来的变长再继续构，得到最大的面积，先得到所有可能的情况，在进行背包——海伦公式 p=(a+b+c)/2;S = sqrt(p*(p-a)*(p-b)*(p-c))；貌似这道题看作边长越大面积越大...背包问题就是一种从最大开始数的情况.

#include<cstdio>

#include<cstring>

#include<cmath>

#include<iostream>

using namespace std;

int dp[1000][1000];

int main(){

    int n,res = 0,mx = -1;

    int a[50];

    scanf("%d",&n);

    for(int i = 1; i <= n; i++){

        scanf("%d", &a[i]);

        res+=a[i];}

        int half = res>>1;

    memset(dp,0,sizeof(dp));

    dp[0][0] = 1;

    for(int i = 1; i <= n;i++)

        for(int j = half; 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]])//得到可能性的边

                     dp[j][k] = 1;

        for(int i = half;i >= 1;i--){

            for(int j = i;j >= 1;j--){

                if(dp[i][j]){

                     int k = res - i - j;

                    if(i+j>k && i+k>j&& k+j>i){

                        double p = 1.0*(j+k+i)/2;

                     int S = (int)(sqrt(p*(p-i)*(p-j)*(p-k))*100);

                    if(S>mx)

                        mx = S;

                    }

                }

            }

        }

        printf("%d",mx);

    return 0;

}

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