hdu 2602(dp)

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

 

 

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 32499    Accepted Submission(s): 13379

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

 

 

 

题意 ：给你一个包的体积，每块骨头的价值和占用的体积，求出可以放入价值最大方案的价值。

分析 ：简单的01背包，纯属模板题，也是我做的第一题背包，就直接贴代码了。

 

 1 #include <cstdio>

 2 #include <cmath>

 3 #include <cstring>

 4 #include <algorithm>

 5 using namespace std;

 6 

 7 //dp[i][j]表示放入第i块骨头并占用j体积的最大价值，

 8 //c[i]表示第i块骨头的体积

 9 //w[i]表示第i块骨头的价值

10 

11 int dp[1111][1111],c[1111],w[1111];

12 int T,N,V;

13 

14 int main ()

15 {

16     int i,j;

17     scanf ("%d",&T);

18     while (T--)

19     {

20         scanf ("%d%d",&N,&V);

21         for (i=1; i<=N; i++)

22         scanf ("%d",&w[i]);

23         for (i=1; i<=N; i++)

24         scanf ("%d",&c[i]);

25         memset(dp, 0, sizeof(dp));

26         for (i=1; i<=N; i++)

27         {

28             for (j=0; j<=V; j++)

29             {

30                 dp[i][j] = dp[i-1][j];   //这里主要考虑j小于c[i]时放不下第i块骨头

31                 if (j >= c[i])

32                 dp[i][j] = max(dp[i-1][j], dp[i-1][j-c[i]]+w[i]);

33             }

34         }

35         printf ("%d\n",dp[N][V]);

36     }

37     return 0;

38 }

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