Triangular Pastures
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 6774 | Accepted: 2199 |
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：
求n条线段构成的三角形的最大面积*100
题解
由于三角形有三个边，并且所有的线段都要用到，于是可以用dp[i][j]表示边长为i,j,sum-i-j的三角形，由于是二维的，j，k边有一个可以取到0，另一个减去a后要大于等于0，最后用海伦公式算面积。
代码：
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
int dp[1005][1005];
int n,sum;
int main()
{
while(~scanf("%d",&n))
{
sum=0;
int a;
memset(dp,0,sizeof(dp));
dp[0][0]=1;
for(int i=0;i<n;i++)
{
scanf("%d",&a);
sum+=a;
for(int j=1000;j>=0;j--)
{
for(int k=1000;k>=0;k--)
{
if((k-a>=0&&dp[j][k-a])||(j-a>=0&&dp[j-a][k]))
{
dp[j][k]=1;
}
}
}
}
int ans=0;
for(int i=1;i<=sum/2;i++)
{
for(int j=1;j<=sum/2;j++)
{
if(dp[i][j]&&(i+j>sum-i-j))
{
double p=sum*1.0/2;
int temp=(int)(sqrt(p*(p-i)*(p-j)*(p-sum+i+j))*100);
ans=max(ans,temp);
}
}
}
if(ans)
printf("%d\n",ans);
else
printf("-1\n");
}
return 0;
}