HDU 2602： Bone Collector

Bone Collector

查看原题

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 67772    Accepted Submission(s): 28288

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 2 ³¹).

 

Sample Input

  
  

   
   1
5 10
1 2 3 4 5
5 4 3 2 1
  
  

 

Sample Output

  
  

   
   14
  
  

 

Author

Teddy

题意：

直白的背包问题了，说的很明白，给定骨头种类数，背包体积。然后又给出每个骨头的价值和体积。

解题思路：

01背包问题。

源代码：

#include<cstdio>
#include<cstring>
const int N=1005;
int max(int a, int b){return a>b?a:b;}
int dp[N];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,V;
        int w[N],v[N];//w价格，v代表体积
        memset(dp,0,sizeof(dp));
        scanf("%d%d",&n,&V);
        for(int i=0;i<n;i++)
            scanf("%d",&w[i]);
        for(int i=0;i<n;i++)
            scanf("%d",&v[i]);
        for(int i=0;i<n;i++)
        {
            for(int j=V;j>=v[i];j--)
                dp[j]=max(dp[j],dp[j-v[i]]+w[i]);
        }
        printf("%d\n",dp[V]);

    }
    return 0;
}