Triangular Pastures （DP）

最新推荐文章于 2019-08-09 18:52:56 发布

原创最新推荐文章于 2019-08-09 18:52:56 发布 · 369 阅读

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探讨了如何使用特定长度的围栏构建最大面积的三角形牧场，采用01背包算法结合海伦公式来解决此问题。

Triangular Pastures

Time Limit: 1000ms

Memory Limit: 30000KB

64-bit integer IO format: %lld Java class name: Main

Submit Status PID: 14277

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

Sample Output

思路：三角形周长一定当三遍差值最小时面积最大，海伦公式与均值不等式

且每段只能选择一次，01 背包

code：

<span style="font-size:18px;">#include<bits/stdc++.h>
int dp[800][800];
using namespace std;
int main()
{
    int i,j,k,n,a[50];
    int tem,ans,mid,sum;
    while(scanf("%d",&n)!=EOF)
    {
        sum=0;
        memset(dp,0,sizeof(dp));
        for(i=0; i<n; i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        mid=sum/2;
        ans=-1;
        dp[0][0]=1;
        for(i=0; i<n; i++)
            for(j=mid; j>=0; j--)
                for(k=mid; k>=0; k--)
                {
                    if((j>=a[i]&&dp[j-a[i]][k])||(k>=a[i]&&dp[j][k-a[i]] ))
                    {
                        dp[j][k]=1;
                        int c=sum-j-k;
                        if(j+k>c&&j+c>k&&k+c>j)
                        {
                            double p=(c+j+k)/2.0;
                                tem=(int)(sqrt(p*(p-k)*(p-j)*(p-c))*100);
                            if(ans<tem)
                                ans=tem;
                        }
                    }
                }
        cout<<ans<<endl;
    }
    return 0;
}</span>