HDU-2602 Bone Collector(01背包）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

 

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

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

 

 

Author

Teddy

 

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

 

Recommend

lcy   |   We have carefully selected several similar problems for you:  1203 2159 2955 1171 2191 

 

01背包（ZeroOnePack）: 有N件物品和一个容量为V的背包。（每种物品均只有一件）第i件物品的费用是c[i]，价值是w[i]。求解将哪些物品装入背包可使价值总和最大。

01背包就是取还是不取的问题

 

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define MAXN 1010
int dp[MAXN][MAXN];  //dp[i][j]表示i个物品，体积为j时的最大价值
int value[MAXN],w[MAXN];
int main()
{
   int t;
   int n,v;
   int i,j;
   scanf("%d",&t);
   while(t--)
   {
     memset(dp,0,sizeof(dp));
     scanf("%d%d",&n,&v);
     for(i=1;i<=n;i++)
     scanf("%d",&value[i]);
     for(i=1;i<=n;i++)
     scanf("%d",&w[i]);
     for(i=1;i<=n;i++)
     for(j=v;j>=0;j--)          //体积是从大到小变化的， for(j=0;j<=v;j++) 也是可以的
     {                          //也就是说这里是有两种情况讨论的
     
         if(j>=w[i])
         dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+value[i]);//不要当前物品是i-1来的， 
         //要也是i-1来的              
         else
         dp[i][j]=dp[i-1][j];   //一开始是没这个物品的,装不下的话也只能这样
     }
     cout<<dp[n][v]<<endl;
   }
    return 0;
}