HDU 2602 Bone Collector(背包模板）

Bone Collector

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

Total Submission(s): 51885    Accepted Submission(s): 21849

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

 

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：

<pre name="code" class="html">#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int V[1010];int value[1010];int dp[1010][1010];int main(){int t,n,v;scanf("%d",&t);while(t--){scanf("%d%d",&n,&v);for(int i=1;i<=n;i++){scanf("%d",&value[i]);}for(int i=1;i<=n;i++){scanf("%d",&V[i]);}memset(dp,0,sizeof(dp));for(int i=1;i<=n;i++){for(int j=0;j<=v;j++){dp[i][j]=dp[i-1][j];//每次都要计算一下不放第i个物品时的价值； if(j>=V[i])//若能放第i个物品； {dp[i][j]=max(dp[i][j],dp[i-1][j-V[i]]+value[i]);} }}printf("%d\n",dp[n][v]);}return 0;}