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

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：

 1  #include<stdio.h>
 2  #include<string.h>
 3  #include<algorithm>
 4  using namespace std;
 5  int w[1100],p[1110];
 6  int f[1110][1110];
 7  int main()
 8  {
 9      int t,n,v;
10      scanf("%d",&t);    
11      while(t--)
12      {
13          scanf("%d%d",&n,&v);        
14          for(int i=1;i<=n;i++)
15          scanf("%d",&p[i]);
16          for(int i=1;i<=n;i++)
17          scanf("%d",&w[i]);        
18          memset(f,0,sizeof(f));                
19          for(int i=1;i<=n;i++)        
20          {
21              for(int j=0;j<=v;j++)
22              {
23                  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 
24                  {
25                      // f[i][j] 表示第 i 件物品在背包容量为 j 时的状态，
26                      //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 
27                      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 
28                            
29                  }
30                  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  
31              }        
32          }
33          printf("%d\n",f[n][v]);                
34      }     
35      return 0;
36  }