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POJ 1948 Triangular Pastures(DP)

2024-07-14 10:36:05   221人阅读

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：给出n条木棍，选择其中一部分拼成三角形，并使得面积最大。

最初的想法设一个四维数组，dp[t][i][j][k]，为当前第t个木棍组成的三角形三边的长度分别为i,j,k是否可能存在（可能存在就是可以由边界推得，但不一定满足要求），然后枚举i,j,k确定是否满足题意并更新最大面积。

如果t是逆序枚举，那么第一维可以省略，因为总长已知，知道i，j，那么k也可以算出。所以简化到dp[i][j].可令i>=j.

如果dp[i-a[t]][j]存在或者dp[i][j-a[t]]存在 ，那么dp[i][j]就可能存在，标记为1.

然后遍历dp[][]，寻找可能存在的dp[i][j]，然后确定是否满足题意并更新答案。

这里为什么要标记之后再遍历，而不是标记之后就直接计算？因为标记之后直接计算可能会出现状态为i或j为0，而且一个状态可能被计算多次。


#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
typedef long long LL;
const int MAX=0x3f3f3f3f;
int area(double a,double b,double c) {
    double s=(a+b+c)/2;
    return 100*sqrt(s*(s-a)*(s-b)*(s-c));
}
int a[45], dp[810][810],n,sum=0;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) {
        scanf("%d",&a[i]);
        sum+=a[i];
    }
    int tmp = sum/2 ,_max=0;
    dp[0][0]=1;
    for(int i=1;i<=n;i++)
        for(int j=tmp;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=tmp;i>=1;i--)
        for(int j=i;j>=1;j--) if( dp[i][j] ) {
            int k=sum-i-j;
            if( i+k<=j || i+j<=k || j+k<=i ) continue;
            int ans = area( i,j,k );
            if( ans > _max ) _max=ans;
        }
    if( _max != 0 ) printf("%d\n",_max);
    else printf("-1\n");
    return 0;
}





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