HDU 2602 Bone Collector 0/1背包

题目链接：HDU 2602 Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 28903    Accepted Submission(s): 11789

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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经典0/1背包问题。

有N件物品和一个容量为V的背包，第i件物品的体积为v[i]，价值为w[i]，求解将哪些物品装入背包才干使总价值最大。

这是最基础的背包问题，每种物品仅仅有一件。且仅仅有两种状态：放与不放。

用子问题定义状态：dp[i][v]表示前i件物品恰好放入一个容量为m的包中可获得的最大价值。

状态转移方程：dp[i][m] = max(dp[i-1][m], dp[i-1][m-v[i]]+w[i]);

可转化为一维情况：dp[m] = max(dp[m], dp[m-v[i]]+w[i]);

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int n, v, dp[1010], value[1010], volume[1010];
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &n, &v);
        for(int i = 1; i <= n; i++)
            scanf("%d", &value[i]);
        for(int i = 1; i <= n; i++)
            scanf("%d", &volume[i]);
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; i++)
            for(int j = v; j >= volume[i]; j--)
                dp[j] = max(dp[j], dp[j-volume[i]]+value[i]);
        printf("%d\n", dp[v]);
    }

    return 0;
}