HDU2602(01背包)

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23977    Accepted Submission(s): 9729

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

//#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<string>
#include<algorithm>
#define MAX 1100
using namespace std;
int dp[MAX][MAX];//行代表个数，列代表体积
int w[MAX], v[MAX], n, W;
int _search(int i, int j){
    if (dp[i][j] >= 0){
        return dp[i][j];
    }
    int res;
    if (i == n){
        res = 0;
    }
    else if (w[i] > j){
        res = _search(i + 1, j);
    }
    else{
        res = max(_search(i + 1, j), _search(i + 1, j - w[i]) + v[i]);
    }
    return dp[i][j] = res;
}
int main()
{
    int T;
    cin >> T;
    while (T--){
        cin >> n >> W;
        for (int i = 0; i < n; i++){
            cin >>v[i];
        }
        for (int i = 0; i < n; i++){
            cin >> w[i];
        }
        memset(dp, -1, sizeof(dp));
        cout << _search(0,W) << endl;
    }
    
    system("pause");
    return 0;
}

做这道题的时候二逼了。其实主要是练习记忆搜索。看来递归并没有我想象中掌握的那么好，革命啥时候能成功啊。。。