题解报告：hdu 2602 Bone Collector（01背包）

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

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

解题思路：简单的01背包（dp）。

AC代码一：（二维数组实现）

 #include<bits/stdc++.h>
 using namespace std;
 const int maxn=;
 int t,n,W,v[maxn],w[maxn],dp[maxn][maxn];
 int main(){
     while(cin>>t){
         while(t--){
             cin>>n>>W;
             for(int i=;i<=n;++i)cin>>v[i];
             for(int i=;i<=n;++i)cin>>w[i];
             memset(dp,,sizeof(dp));
             for(int i=;i<=n;++i){
                 for(int j=;j<=W;++j){
                     if(j<w[i])dp[i][j]=dp[i-][j];//无法挑选这个物品
                     else dp[i][j]=max(dp[i-][j],dp[i-][j-w[i]]+v[i]);//拿和不拿的两种情况都试一下
                 }
             }
             cout<<dp[n][W]<<endl;
         }
     }
     return ;
 }

AC代码二：（一维数组实现）

 #include<bits/stdc++.h>
 using namespace std;
 int value[],weight[],dp[];//dp数组始终记录当前体积的最大价值
 int main()
 {
     int T,N,V;
     cin>>T;
     while(T--){
         cin>>N>>V;
         for(int i=;i<N;i++)cin>>value[i];//输入价值
         for(int i=;i<N;i++)cin>>weight[i];//输入体积
         memset(dp,,sizeof(dp));//初始化
         for(int i=;i<N;i++){    //个数
             for(int j=V;j>=weight[i];j--) //01背包
                 dp[j]=max(dp[j],dp[j-weight[i]]+value[i]); //比较放入i物体后的价值与不放之前的价值，记录大的值
         }
         cout<<dp[V]<<endl;//输出总体积的最大价值
     }
     return ;
 }