Charm Bracelet poj 01 背包

最新推荐文章于 2022-04-25 19:27:35 发布

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本文介绍了一个经典的背包问题案例，通过动态规划算法解决如何在重量限制内选择最优组合的问题。使用一维数组进行优化，避免了二维数组带来的超时问题。

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Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17400		Accepted: 7901

Description

Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight W_i (1 ≤ W_i ≤ 400), a 'desirability' factor D_i (1 ≤ D_i ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).

Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: W_i and D_i

Output

* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

Sample Input

Sample Output

Source

这道题本身并没有什么难度，只是基础题~ 拿来复习用的~

#include<iostream>
using namespace std;
int max(int a,int b)
{
	return a>b?a:b;
}
int main()
{
	int num,totalweight,weight[12880],value[12880];
	int f[12880];
	cin>>num>>totalweight;
	for(int i=0;i<num;i++)
	{
		cin>>weight[i]>>value[i];
	}
	for(int i=0;i<num;i++)
	{
		 for(int v=totalweight;v>=weight[i];v--)
		 {
 
 			f[v]=max(f[v],f[v-weight[i]]+value[i]);
 		}
	}
	cout<<f[totalweight]<<endl;
	return 0;
}

后来用二维数组写一下~ 超时~

code:

#include<iostream>
#include<string.h>
using namespace std;
int v[12881]={0};
int w[12881]={0};
int dp[12881][12881];
int main()
{
    int n,V;
    while (cin>>n>>V)
    {
        for (int i=1;i<=n;i++)
        {
            cin>>v[i]>>w[i];           
        }

        memset(dp,0,sizeof(dp));  
            
         for(int i=1;i<=n;i++)
            for(int j=0;j<=V;j++) 
		   {
                if(v[i]<=j )
                    dp[i][j]=max(dp[i-1][j-v[i]]+w[i],dp[i-1][j]); 
				else
				  dp[i][j]=dp[i-1][j];  //这一步一定要加上~             
            }

          cout<<"---------------->(x轴) 体积"<<endl;;
           for (int i=1;i<=n;i++)
           {
           	cout<<"|";
            for (int j=0;j<=V;j++)
            {
                cout<<dp[i][j]<<" ";
            }
            cout<<endl;
        }
        cout<<"|"<<endl;
        cout<<"Y轴 n"<<endl; 
        
        cout<<dp[n][V]<<endl;
    }
    return 0;

}

二维数组要注意的是那个

else

dp[i][j]=dp[i-1][j];

由于在上一步中如果超了体积应直接赋值给dp[i][j]

不加else

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

4 6
1 4
2 6
3 12
2 7
---------------->(x轴) 体积
|0 4 4 4 4 4 4
|0 0 6 10 10 10 10
|0 0 0 12 12 18 22
|0 0 7 12 12 19 22
|
Y轴 n
22

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

正确的：

4 6
1 4
2 6
3 12
2 7
---------------->(x轴) 体积
|0 4 4  4  4  4  4
|0 4 6 10 10 10 10
|0 4 6 12 16 18 22
|0 4 7 12 16 19 23
|
Y轴 n
23

也就是说在给出的v值如果有相同值的时候 不加else 得出的答案会有错误~