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 ?

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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.

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Output

One integer per line representing the maximum of the total value (this number will be less than 2³¹).

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Sample Input

1 5 10 1 2 3 4 5 5 4 3 2 1

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Sample Output

14

題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=2602

題意：給出n件物品的重量和價值，放進一個容量為v的背包，使背包里的價值最大。

0-1背包的模板題，可以有兩種解法。

一維數組：

#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]); //輸入物品重量 for(int i=1;i<=n;i++)scanf("%d",&p[i]); //輸入物品價值       memset(f,0,sizeof(f));for(int i=1;i<=n;i++)  {for(int j=v;j>=w[i];j--) //這個循環保證了放進去的物品重量不會超過背包所能容納的重量 
            {if(f[j] < f[j-w[i]] + p[i])  // 如果當前所擁有價值  小于 加上這件物品時創造的價值就更新 f[j]= f[j-w[i]] + p[i]; //   f[j] 表示背包重量為 j 時背包里的最大價值，//  所以f[ j - w[i] ] 表示放進這件物品時的狀態（因為放進該件物品后容量就減少了）
            }}    printf("%d\n",f[v]);}        return 0;
}

二維數組：

 #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) // 這件物品的重量小于當前的容量，也就是說放的進背包 
                 {
                     // f[i][j] 表示第 i 件物品在背包容量為 j 時的狀態，
                     //所以 f[i-1][j] 表示背包在上一次容量為 j 時候的狀態，也就是沒放這件物品的時候 
                     f[i][j]=max(f[i-1][j],f[i-1][j-w[i]]+p[i]);// 比較 沒放進去之前 和放進該物品后 的價值，取最大 
                           
                 }
                 else f[i][j]=f[i-1][j];  // 如果不能放進該物品，則取上一次的狀態  
             }        
         }
         printf("%d\n",f[n][v]);                
     }     
     return 0;
 }

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渣渣一枚，如果有什么不對的地方，還請各位大神批評指正~ ?(^_^)