HDU-2602 Bone Collector

http://acm.hdu.edu.cn/showproblem.php?pid=2602

01背包：用二维数组实现。

c[n][m]表示n种物品，背包容量为m的最大价值。

状态方程为:f(n,m)=max{f(n-1,m), f(n-1,m-w[n])+P(n,m)}这就是书本上写的动态规划方程.

 

             Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 24513    Accepted Submission(s): 9911

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<cstring>
#include<cstdio>
using namespace std;
int c[1010][1010];
int p[2000],w[2000];
int main()
{
    int t,n,m,i,j;
    cin>>t;
    while(t--)
   {
        memset(p,0,sizeof(p));
        memset(w,0,sizeof(w));
       memset(c,0,sizeof(c));
        cin>>n>>m;
      for(i=1;i<=n;i++)
        cin>>p[i];
      for(i=1;i<=n;i++)
        cin>>w[i];
    for(i=1;i<n+1;i++)
    for(j=0;j<m+1;j++)
    {
        if(w[i]<=j)
        {
             if(p[i]+c[i-1][j-w[i]]>c[i-1][j])
                 c[i][j]=p[i]+c[i-1][j-w[i]];
             else
                 c[i][j]=c[i-1][j];
        }else
             c[i][j]=c[i-1][j];
     }
      printf("%d\n",c[n][m]);

    }
    return 0;
}