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 2 ³¹).

 

 

Sample Input

1 5 10 1 2 3 4 5 5 4 3 2 1

 

 

Sample Output

14

 

 

Author

Teddy

 

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

 

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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：