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01背包N

时间:2015-08-11 17:51:14      阅读:125      评论:0      收藏:0      [点我收藏+]

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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
 
 
 
思路:
       状态转移方程:val[j]=max{val[j],val[j-weight[i]]+p[i]}
       val[j]表示:前i个物体在容量为vol最大的装载量
 
源代码:
 
 1 #include<iostream>
 2 #include<algorithm>
 3 #include<string>
 4 #include<cstring>
 5 #include<cmath>
 6 using namespace std;
 7 #define maxn 1000+10
 8 int val[maxn], weight[maxn], p[maxn];
 9 int main()
10 {
11     int T;
12     cin >> T;
13     while (T--)
14     {
15         int n, vol;
16         cin >> n >> vol;
17         for (int i = 1; i <= n; i++)
18         {
19             cin >> p[i];
20         }
21         for (int j = 1; j <= n; j++)
22         {
23             cin >> weight[j];
24         }
25         for (int j = 0; j <= vol; j++)
26         {
27             val[j] = 0;
28         }
29         for (int i = 1; i <= n; i++)  
30         {
31             for (int j = vol; j >= weight[i]; j--)
32             {
33                 val[j] = max(val[j], val[j - weight[i]] + p[i]);
34 
35             }
36         }
37         cout << val[vol] << endl;  
38 
39     }
40        return 0;
41 
42 }

 

心得:

 背包入门~~~~理解一个方程还是挺费劲的,/(ㄒoㄒ)/~~不过终于比之前好多了。。。。给自己一个赞!不过还是觉得好神奇

                                      (*^__^*) 嘻嘻……         

 

 

 
 
 

01背包N

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原文地址:http://www.cnblogs.com/Lynn0814/p/4721406.html

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