Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 60469 Accepted Submission(s): 25209
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
Author
Teddy
Source
HDU 1st “Vegetable-Birds Cup” Programming Open Contest
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2602
分析:01背包裸题,注意格式输入输出就行了,我就是输入格式写错了,找错误找了一个小时,QAQ
下面给出AC代码:
1 #include <bits/stdc++.h>
2 using namespace std;
3 int main()
4 {
5 int n;
6 int v[1010],dp[1010],d[1010];//v代表体积,d代表谷歌值
7 while(scanf("%d",&n)!=EOF)
8 {
9 while(n--)
10 {
11 memset(v,0,sizeof(v));
12 memset(dp,0,sizeof(dp));
13 memset(d,0,sizeof(d));
14 int x,y;
15 cin>>x>>y;
16 for(int i=1;i<=x;i++)
17 cin>>d[i];
18 for(int i=1;i<=x;i++)
19 cin>>v[i];
20 for(int i=1;i<=x;i++)//01背包主函数
21 {
22 for(int j=y;j>=v[i];j--)
23 {
24 dp[j]=max(dp[j],dp[j-v[i]]+d[i]);
25 }
26 }
27 printf("%d\n",dp[y]);
28 }
29 }
30 return 0;
31 }