HDU 2602 Bone Collector(01背包裸题)

Bone Collector

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

题意：一个叫做Bone Collector的男的有一个包，往包里放东西，使得其价值最大。

输入：注意是先输入的是价值，后是体积。

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：

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    int v[1010],dp[1010],d[1010];//v代表体积，d代表谷歌值
    while(scanf("%d",&n)!=EOF)
    {
        while(n--)
        {
           memset(v,0,sizeof(v));
           memset(dp,0,sizeof(dp));
           memset(d,0,sizeof(d));
           int x,y;
           cin>>x>>y;
           for(int i=1;i<=x;i++)
              cin>>d[i];
           for(int i=1;i<=x;i++)
              cin>>v[i];
           for(int i=1;i<=x;i++)//01背包主函数
           {
               for(int j=y;j>=v[i];j--)
               {
                dp[j]=max(dp[j],dp[j-v[i]]+d[i]);
               }
           }
           printf("%d
",dp[y]);
        }
    }
    return 0;
}