Bone Collector HDU - 2602

最新推荐文章于 2020-02-07 20:27:03 发布

转载最新推荐文章于 2020-02-07 20:27:03 发布 · 114 阅读

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原文链接：http://www.cnblogs.com/zgglj-com/p/6810482.html

本文介绍了一道经典的01背包问题，通过一个喜欢收集骨头的人的故事背景，详细解析了如何利用动态规划算法来求解在有限背包容量下所能获得的最大价值。

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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 ?

InputThe 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.OutputOne integer per line representing the maximum of the total value (this number will be less than 2 ³¹).Sample Input

Sample Output

14

裸的01背包

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<cstring>
 5 using namespace std;
 6 
 7 int main()
 8 {   int t,n,V;
 9     cin>>t;
10     while(t--){
11         scanf("%d%d",&n,&V);
12         int va[1005],vo[1005],dp[1005];
13         for(int i=0;i<n;i++) scanf("%d",&va[i]);
14         for(int i=0;i<n;i++) scanf("%d",&vo[i]);
15         
16         memset(dp,0,sizeof(dp));
17         for(int i=0;i<n;i++){
18             for(int j=V;j>=vo[i];j--)
19                 dp[j]=max(dp[j],dp[j-vo[i]]+va[i]);
20         }
21         printf("%d\n",dp[V]);
22     }
23 }

转载于:https://www.cnblogs.com/zgglj-com/p/6810482.html