HDU 2602 (0-1背包)

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 35815    Accepted Submission(s): 14753

Problem Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

 

 

Input

The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

 

 

Output

One integer per line representing the maximum of the total value (this number will be less than 2³¹).

 

 

Sample Input

1 5 10 1 2 3 4 5 5 4 3 2 1

 

 

Sample Output

14

 

 

Author

Teddy

 

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

 

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题目大意：

　　就是说，给你一个N和V和N个物品的v[i]和w[i]，让你求出，并且每次向背包中放入一个物品，让你求出最多能放多少个物品(物品的体积<=V)所能带来的最大的价值。

解题思路：

　　直接走0-1背包的模型：

　　　　定义状态：dp[i+1][j]表示，从前i个物品中选出来总重不超过j的物品所能带来的最大价值.

　　　　初始状态：dp[0][j]==0.

　　　　状态转移方程: dp[i+1][j] = dp[i][j] ,当j < w[i]时。

　　　　　　　　　　  dp[i+1][j] = max ( dp[i][j],dp[i][j-w[i]]+v[i]); 其他

# include<cstdio>
# include<iostream>

using namespace std;

const int MAX = 1000+4;

int dp[MAX][MAX];
int w[MAX];
int v[MAX];


int main(void)
{
    int t;cin>>t;
    while ( t-- )
    {
        int n,vv;
        cin>>n>>vv;
        for ( int i = 0;i < n;i++ )
        {
            cin>>v[i];
        }
        for ( int i = 0;i < n;i++ )
        {
            cin>>w[i];
        }
        for ( int i = 0;i < n;i++ )
        {
            for ( int j = 0;j <= vv;j++ )
            {
                dp[0][j] = 0;
                if ( j < w[i] )
                {
                    dp[i+1][j] = dp[i][j];
                }
                else
                {
                    dp[i+1][j] = max(dp[i][j],dp[i][j-w[i]]+v[i] );
                }
            }
        }
        cout<<dp[n][vv]<<endl;
    }


    return 0;
}

 

代码：