HDOJ 2602 Bone Collector 0-1背包问题的最简形

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 60848    Accepted Submission(s): 25366

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
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <cctype>
#define N 1010
#define Max 0xffff
int dp[N][N];
//表示取前i种物品，使它们的总体积不超过j的最优取法取得的价值总和
using namespace std;
int main()
{
    int cas,n,V;
    cin>>cas;
    while(cas--)
    {
        memset(dp,0,sizeof(dp));
        int d[N] {0},w[N] {0};
        /*用两个数组分别存储价值和体积*/
        cin>>n>>V;
        for(int i=1; i<=n; ++i)
            cin>>d[i];
        for(int i=1; i<=n; i++)
            cin>>w[i];
        for(int i=1; i<=n; i++)
        {
            for(int j=0; j<=V; j++)
            {
                if(w[i]<=j)
                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+d[i]);
                    //对于每一件物品，有放或者不放两种选择(如果能放下)
                else
                    dp[i][j]=dp[i-1][j];
            }
        }
        cout<<dp[n][V]<<endl;
    }
    return 0;
}