HDU2602—Bone Collector（背包问题）

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

01背包问题，这种背包特点是：每种物品仅有一件，可以选择放或不放。
用子问题定义状态：即dp[i][v]表示前i件物品恰放入一个容量为v的背包可以获得的最大价值。
则其状态转移方程便是：
dp[i][v]=max{dp[i-1][v],dp[i-1][v-cost[i]]+value[i]}

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <string>
#define MAXN 1000010
using namespace std;
int max(int a,int b)
{
if(a>b)
return a;
else
return b;
}
int main()
{
int t,n,v,i,j,c[1010],w[1010],dp[1010];
scanf("%d",&t);
while(t--)
{
memset(dp,0,sizeof(dp));
scanf("%d%d",&n,&v);
for(i=0; i<n; ++i)
scanf("%d",&c[i]);
for(i=0; i<n; ++i)
scanf("%d",&w[i]);
for(i=0; i<n; ++i)
for(j=v; j>=0; --j)
{
if(j>=w[i])
dp[j]=max(dp[j],dp[j-w[i]]+c[i]);
}
cout<<dp[v]<<endl;
}
return 0;
}