Triangular Pastures
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 6783 | Accepted: 2201 |
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]
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<cstdlib>
using namespace std;
int n,a[50],dp[810][810],sum,p,ans=-1;
int main()
{
memset(dp,0,sizeof(dp));
scanf("%d",&n);
dp[0][0]=1;
for(int i=1;i<=n;i++)
scanf("%d",&a[i]),sum+=a[i];
for(int i=1;i<=n;i++)
{
for(int j=sum/2;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]]))
{
dp[j][k]=1;
}
}
}
}
for(int i=sum/2;i>=1;i--)
{
for(int j=i;j>=1;j--)
{
if(dp[i][j])
{
int k=sum-i-j;
double p=sum*1.0/2;
if(i+j>k&&i+k>j&k+i>j)
{
int temp=(int)(sqrt(p*(p-i)*(p-j)*(p-k))*100);
if(temp>ans)
ans=temp;
}
}
}
}
printf("%d\n",ans);
return 0;
}