二维费用背包--poj1948

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最新推荐文章于 2025-03-23 16:42:09 发布

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分类专栏：背包动态规划

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Triangular Pastures

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 5789		Accepted: 1855

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

Sample Output

Hint

[which is 100x the area of an equilateral triangle with side length 4]

首先说自己没想出来，不知道怎么把他抽象成二维背包，看了大神们的解释才明白了些。
把边的长度看成费用，

三角形中的某条边 i 看作第一维费用，另一条边看作是第二维费用 j，背包容量为总长的一半（三角形任意一条边都不可能比周长的一半大）则第三边为总长度sum-i-j;
找出所有满足的i,j，枚举求出最大面积即可。其中面积公式为{ 1.t=sum/2; 2.area=sqrt(t*(t-i)*(t-j)*(t-(sum-i-j));}
状体转移方程为   dp[ k ][ i ][ j ] = dp[ k-1 ][ i-sticks[k] ][ j ] | dp[ k-1 ][ i ][ j-sticks[k] ]; 意思是取第k条棍子的时候，组成的两条边长度分别为 i 和 j 。

AC代码如下：

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
using namespace std;
int f[50];
int dp[805][805];
int main()
{
    //freopen("in.txt","r",stdin);
    int n;
    while(cin>>n)
    {
        int sum=0;
        for(int i=1; i<=n; i++)
        {
            cin>>f[i];
            sum+=f[i];
        }
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;
        int mid=sum/2;
        for(int i=1; i<=n; i++)
            for(int j=mid; j>=0; j--)
                for(int k=mid; k>=0; k--)
                    if((j>=f[i]&&dp[j-f[i]][k])||(k>=f[i]&&dp[j][k-f[i]]))//只要j-f[i]或者k-f[i]中存在里挑，就是可行的
                        {dp[j][k]=1;}
       int ans=-1;
        for(int i=mid; i>0; i--)
            for(int j=mid; j>0; j--)
                if(dp[i][j])
                {
                    int q=sum-i-j;
                    if(i+j>q&&i+q>j&&j+q>i)
                    {
                        double p=sum/2.0;
                        double x=sqrt(p*(p-i)*(p-j)*(p-q));//海伦公式
                        if(x*100>ans)
                              ans=x*100;
                    }
                }
        cout<<ans<<endl;
            //printf("%d\n",ans);
    }
    return 0;
}