HDU 2602 (0-1背包)

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 35815    Accepted Submission(s): 14753


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
 

 

Author
Teddy
 

 

Source
 

 

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题目大意:

  就是说,给你一个N和V和N个物品的v[i]和w[i],让你求出,并且每次向背包中放入一个物品,让你求出最多能放多少个物品(物品的体积<=V)所能带来的最大的价值。

解题思路:

  直接走0-1背包的模型:

    定义状态:dp[i+1][j]表示,从前i个物品中选出来总重不超过j的物品所能带来的最大价值.

    初始状态:dp[0][j]==0.

    状态转移方程: dp[i+1][j] = dp[i][j] ,当j < w[i]时。

            dp[i+1][j] = max ( dp[i][j],dp[i][j-w[i]]+v[i]); 其他

 1 # include<cstdio>
 2 # include<iostream>
 3 
 4 using namespace std;
 5 
 6 const int MAX = 1000+4;
 7 
 8 int dp[MAX][MAX];
 9 int w[MAX];
10 int v[MAX];
11 
12 
13 int main(void)
14 {
15     int t;cin>>t;
16     while ( t-- )
17     {
18         int n,vv;
19         cin>>n>>vv;
20         for ( int i = 0;i < n;i++ )
21         {
22             cin>>v[i];
23         }
24         for ( int i = 0;i < n;i++ )
25         {
26             cin>>w[i];
27         }
28         for ( int i = 0;i < n;i++ )
29         {
30             for ( int j = 0;j <= vv;j++ )
31             {
32                 dp[0][j] = 0;
33                 if ( j < w[i] )
34                 {
35                     dp[i+1][j] = dp[i][j];
36                 }
37                 else
38                 {
39                     dp[i+1][j] = max(dp[i][j],dp[i][j-w[i]]+v[i] );
40                 }
41             }
42         }
43         cout<<dp[n][vv]<<endl;
44     }
45 
46 
47     return 0;
48 }

 

代码:

posted @ 2015-04-15 18:01  BYYB_0506  阅读(142)  评论(0编辑  收藏  举报