HDU 2602 Bone Collector(01背包裸题)

时间：2023-03-31 22:23:20

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 ?HDU 2602 Bone Collector(01背包裸题)
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
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[],dp[],d[];//v代表体积，d代表谷歌值

     while(scanf("%d",&n)!=EOF)

     {

         while(n--)

         {

            memset(v,,sizeof(v));

            memset(dp,,sizeof(dp));

            memset(d,,sizeof(d));

            int x,y;

            cin>>x>>y;

            for(int i=;i<=x;i++)

               cin>>d[i];

            for(int i=;i<=x;i++)

               cin>>v[i];

            for(int i=;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\n",dp[y]);

         }

     }

     return ;

 }

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