HDOJ-2602 Bone Collector [DP-01背包问题]

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12502    Accepted Submission(s): 4874

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

 

 

 

code:

#include<iostream>
using namespace std;

int main()
{
    int N,V;
    int n[1001],v[1001],dp[1001];
    int T;
    int i,j;
    while(~scanf("%d",&T))
    {
        while(T--)
        {
            scanf("%d%d",&N,&V);
            for(i=1;i<=N;i++)
                scanf("%d",&v[i]);
            for(i=1;i<=N;i++)
                scanf("%d",&n[i]);
 //------------------------------------------------------------------------------------------------------------- 
            memset(dp,0,sizeof(dp));
            for(i=1;i<=N;i++)
                for(j=V;j>=n[i];j--)
                {
                    if(dp[j-n[i]]+v[i]>dp[j])
                        dp[j]=dp[j-n[i]]+v[i];
                }
 //------------------------------------------------------------------------------------------------------------- 
            printf("%d\n",dp[V]);
        }
    }
    return 0;
} 

 

 

自己调试结果：

最大容量V	物品个数N	Bone Collector
10	5
物品大小n[i]	物品价值v[i]	编号	DP	10	9	8	7	6	5	4	3	2	1
5	1	1	1	1	1	1	1	1	1	0	0	0	0
4	2	2	2	3	3	2	2	2	2	2	0	0	0
3	3	3	3	6	5	5	5	3	3	3	3	0	0
2	4	4	4	9	9	7	7	7	7	7	4	4	0
1	5	5	5	14	12	12	12	12	12	9	9	5	5