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Bone Collector(ZeroOnebag)

时间:2015-08-02 11:41:34      阅读:91      评论:0      收藏:0      [点我收藏+]

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Bone Collector

 

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 ?
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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
 模板题,找到状态转移方程,使用一维01背包,对于背包的体积要逆序遍历,这样是的右边的max(f[j],f[j-v[i]]+val[i])能够表示为上一层的状态,而对于当j<v[i时,默认f[j]不变,是上一层的值
 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstring>
 4 using namespace std;
 5 int main()
 6 {
 7     int T,V;
 8     int v[1001],val[1001];
 9     int f[1001];
10     int i,N,j;
11     cin>>T;
12     while(T--)
13     {
14         memset(f,0,sizeof(f));
15         cin>>N>>V;
16         for(i=1;i<=N;i++)
17             cin>>val[i];
18         for(i=1;i<=N;i++)
19             cin>>v[i];
20         for(i=1;i<=N;i++)
21         {
22             for(j=V;j>=v[i];j--)
23                 f[j]=max(f[j],f[j-v[i]]+val[i]);
24         }
25         cout<<f[V]<<endl;
26     }
27 }

 

Bone Collector(ZeroOnebag)

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原文地址:http://www.cnblogs.com/a1225234/p/4695347.html

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