HDU 2602 Bone Collector（dp）

最新推荐文章于 2022-10-28 13:05:08 发布

原创最新推荐文章于 2022-10-28 13:05:08 发布 · 339 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#dp

ACM/ICPC题解专栏收录该内容

166 篇文章

订阅专栏

本文介绍了一个经典的01背包问题实例，通过一个骨骸收集者的案例来解释如何使用动态规划解决此类问题。文章详细展示了输入输出样例及AC代码实现过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 57246 Accepted Submission(s): 23894

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

Sample Output

14





 01背包问题 定义一个数组dp[i][j]  i 表示第几个物体 j表示背包的体积  背包的体积从0开始到背包本身体积v结束 每次求出	  在该体积下的最优解 在所有物体均讨论完毕后  得到整体最优解


AC代码：
 #include <stdio.h>  
#include <string.h>  
int dp[1005][1005];  
  
  
int max(int a,int b){  
    return a>b?a:b;   
}  
  
      
int main (){  
    int T;  
    int N,V;  
    int weight[1005],value[1005];  
  
    scanf ("%d",&T);  
      
    while (T--){  
          
        scanf ("%d%d",&N,&V);  
          
        for (int i=1;i<=N;i++){  
            scanf ("%d",&value[i]);//价值   
        }  
          
        for (int i=1;i<=N;i++){  
            scanf ("%d",&weight[i]);//重量   
        }  
          
        memset(dp,0,sizeof(dp));//初始数组dp为零   
          
           
        for (int i=1;i<=N;i++)  
        {  
            for (int j=0;j<=V;j++)  
            {  
                if (weight[i]<=j)  
                {  
                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-weight[i]]+value[i]);//判断 是 牺牲这一部分空间加上该物体的价值 还是 不牺牲空间选取上一个最优解   
                }  
                else{  
                    dp[i][j]=dp[i-1][j];  
                }  
            }   
        }   
        printf ("%d\n",dp[N][V]);//最后得到整体最优解   
    }  
    return 0;  
}