【hdu 2602 Bone Collector（动态规划、01背包）】

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12318    Accepted Submission(s): 4787

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

 

Recommend

lcy

 

 

// Project name : 2602 ( Bone Collector ) 
// File name    : main.cpp
// Author       : Izumu
// Date & Time  : Sat Jul 14 10:03:36 2012


#include <iostream>
#include <stdio.h>
#include <string>
#include <cmath>
#include <algorithm>
using namespace std;

#define MAXN 1010

int value[MAXN];
int cost[MAXN];

int dp[MAXN][MAXN];

int n, v;

int max(int a, int b)
{
    return (a > b)? a: b;
}

void DynamicProgramming()
{
    // init for dp[][]
    for (int i = 0; i <= n; i++)
    {
        for (int j = 0; j <= v; j++)
        {
            dp[i][j] = 0;
        }
    }

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

        DynamicProgramming();

        cout << dp[n][v] << endl;
    }
    return 0;
}

// end 
// ism