HDU2602 （0-1背包）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 39259    Accepted Submission(s): 16261

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

 

 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2602

 

题意：给出n件物品的重量和价值，放进一个容量为v的背包，使背包里的价值最大。

 

0-1背包的模板题，可以有两种解法。

 

一维数组：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int  w[1100],p[1110];
int f[1110];
int main() 
{
    int t,n,v;
    scanf("%d",&t);    
    while(t--)
    {
        scanf("%d%d",&n,&v);
        for(int i=1;i<=n;i++)
        scanf("%d",&w[i]); //输入物品重量 
        for(int i=1;i<=n;i++)
        scanf("%d",&p[i]); //输入物品价值       
        memset(f,0,sizeof(f));
        for(int i=1;i<=n;i++)  
        {
            for(int j=v;j>=w[i];j--) //这个循环保证了放进去的物品重量不会超过背包所能容纳的重量 
            {
                if(f[j] < f[j-w[i]] + p[i])  // 如果当前所拥有价值  小于 加上这件物品时创造的价值就更新 
                f[j]= f[j-w[i]] + p[i]; 
                //   f[j] 表示背包重量为 j 时背包里的最大价值，
                //  所以f[ j - w[i] ] 表示放进这件物品时的状态（因为放进该件物品后容量就减少了）
            }
        }    
        printf("%d\n",f[v]);
    }        
    return 0;
}

二维数组：

 #include<stdio.h>
 #include<string.h>
 #include<algorithm>
 using namespace std;
 int w[1100],p[1110];
 int f[1110][1110];
 int main()
 {
     int t,n,v;
     scanf("%d",&t);    
     while(t--)
     {
         scanf("%d%d",&n,&v);        
         for(int i=1;i<=n;i++)
         scanf("%d",&p[i]);
         for(int i=1;i<=n;i++)
         scanf("%d",&w[i]);        
         memset(f,0,sizeof(f));                
         for(int i=1;i<=n;i++)        
         {
             for(int j=0;j<=v;j++)
             {
                 if(w[i]<=j) // 这件物品的重量小于当前的容量，也就是说放的进背包 
                 {
                     // f[i][j] 表示第 i 件物品在背包容量为 j 时的状态，
                     //所以 f[i-1][j] 表示背包在上一次容量为 j 时候的状态，也就是没放这件物品的时候 
                     f[i][j]=max(f[i-1][j],f[i-1][j-w[i]]+p[i]);// 比较 没放进去之前 和放进该物品后 的价值，取最大 
                           
                 }
                 else f[i][j]=f[i-1][j];  // 如果不能放进该物品，则取上一次的状态  
             }        
         }
         printf("%d\n",f[n][v]);                
     }     
     return 0;
 }

 

渣渣一枚，如果有什么不对的地方，还请各位大神批评指正~  (^_^)