Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 52296 Accepted Submission(s): 22040

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

5 10

1 2 3 4 5

5 4 3 2 1

Sample Output

裸的01背包

 1 //2016.9.6
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5
 6 using namespace std;
 7
 8 const int N = 1005;
 9 int dp[N], c[N], w[N];
10
11 int main()
12 {
13     int T, n, v;
14     cin>>T;
15     while(T--)
16     {
17         scanf("%d%d", &n, &v);
18         memset(dp, 0, sizeof(dp));
19         for(int i = 0; i < n; i++)
20               scanf("%d", &w[i]);
21         for(int i = 0; i < n; i++)
22               scanf("%d", &c[i]);
23         for(int i = 0; i < n; i++)
24               for(int j = v; j >= c[i]; j--)
25                   dp[j] = max(dp[j], dp[j-c[i]]+w[i]);
26         printf("%d\n", dp[v]);
27     }
28
29     return 0;
30 }

转载于:https://www.cnblogs.com/Penn000/p/5849076.html

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