hdu 2602Bone Collector（0/1背包）

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

这是简单的0/1背包，用一维的比较快。0/1背包，每个物品只能使用一次。

这是最基础的背包问题，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f[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：

f[i][v]=max{f[i-1][v],f[i-1][v-c[i]]+w[i]}

 

代码：

 

#include<iostream>#include<cmath>using namespace std;int dp[1005];int w[1005],c[1005];int main(){    int i,j,n,v,t;    scanf("%d",&t);    while(t--)    {      scanf("%d%d",&n,&v);      memset(dp,0,sizeof(dp));      for(i=1;i<=n;i++)        scanf("%d",&w[i]);      for(i=1;i<=n;i++)        scanf("%d",&c[i]);      for(i=1;i<=n;i++)      {        for(j=v;j>=c[i];j--)        {                     dp[j]=max(dp[j],dp[j-c[i]]+w[i]);        }      }      printf("%d\n",dp[v]);    }    return 0;}