HDU 2602 DP大水题 01背包

除了LCS以外过的第一个DP 而且是第一个用一维数组的 DP  题目及AC代码如下

基本的01背包问题的状态转移方程:

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 25275    Accepted Submission(s): 10244


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
1
5 10
1 2 3 4 5
5 4 3 2 1
 

Sample Output
14
 
 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #include<string.h>
 4 
 5 const int max_n=1010;
 6 
 7 int max(int a,int b)
 8 {
 9     return a>b?a:b;
10 }
11 
12 int main(void)
13 {
14 
15     int cases;
16     int v[max_n];
17     int w[max_n];
18     int dp[max_n];
19     int N,V;
20     scanf("%d",&cases);
21     while(cases--)
22     {
23         memset(dp,0,sizeof(dp));
24         scanf("%d%d",&N,&V);
25         for(int i=0;i<N;i++)
26         {
27             scanf("%d",&w[i]);              //value
28         }
29         for(int i=0;i<N;i++)
30         {
31             scanf("%d",&v[i]);              //volume
32         }
33         for(int i=0;i<N;i++)
34             for(int vi=V;vi>=v[i];vi--)
35                 dp[vi]=max(dp[vi],dp[vi-v[i]]+w[i]);
36 
37         printf("%d
",dp[V]);
38     }
39     return 0;
40 }
 
 
原文地址:https://www.cnblogs.com/VOID-133/p/3631331.html