HDOJ2602（DP0-1背包）

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

Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K(Java/Others)
Total Submission(s): 11339 Accepted Submission(s):4370

Problem Description

Many years ago , in Teddy’s hometown there was a man who wascalled “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 histrip of collecting there are a lot of bones , obviously , differentbone has different value and different volume, now given the eachbone’s value along his trip , can you calculate out the maximum ofthe total value the bone collector can get ?

 

Input

The first line contain a integer T , the number ofcases.
Followed by T cases , each case three lines , the first linecontain two integer N , V, (N <= 1000 , V<= 1000 )representing the number of bones and thevolume of his bag. And the second line contain N integersrepresenting the value of each bone. The third line contain Nintegers representing the volume of each bone.

 

Output

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

 

Sample Input

1 5 10 1 23 4 5 5 4 3 2 1

 

Sample Output

14

 

Author

Teddy

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

Recommend

lcy

具体解法我就不说了。。不会的看背包九讲（重温这题就是为了弄解法1的）不过有三个小问题要注意。。f数组一定要初始化，要不然WA。。还有就是return f[m]与return f[n];居然都能AC。。这个我暂时不理解。。。不过还是建议returnf[m]..三.是由于题目并没有要求一定要把背包装满所以数组初始化的时候应该将f[0..V]全部设为0。

解法1：15MS292K574 B

#include<stdio.h>
#include<string.h>
int n,m;
int f[2010],weight[2010],cost[2010];
int max(intx,int y)
{
   return x>y?x:y;
}
int zeroonepack(intcost,int weight)
{
   int i;
   for(i=m;i>=cost;i--)
       f[i]=max(f[i],f[i-cost]+weight);
   return f[n];
}
int main()
{
   int t;
   scanf("%d",&t);
   while(t--)
   {
       scanf("%d%d",&n,&m);
       int i;
       for(i=1;i<=n;i++)
           scanf("%d",&weight[i]);
       for(i=1;i<=n;i++)
           scanf("%d",&cost[i]);
       memset(f,0,sizeof(f));//这一步一定不可少
       for(i=1;i<=n;i++)
           zeroonepack(cost[i],weight[i]);
       printf("%d\n",f[m]);
   }
   return 0;
}

 

解法215MS264K484 B

#include<stdio.h>
#include<string.h>
intf[2010]={0},c[1010],w[1010];
int max(intx,inty)
{
   if(x>y)returnx;
    elsereturn y;
}
int main()
{
   int T;
   scanf("%d",&T);
   while(T--)
   {
       int n,v,i,j;
       scanf("%d%d",&n,&v);
       for(i=1;i<=n;i++)
           scanf("%d",&w[i]);
       for(i=1;i<=n;i++)
           scanf("%d",&c[i]);
       memset(f,0,sizeof(f));
       for(i=1;i<=n;i++)
       {
           for(j=v;j>=c[i];j--)
               f[j]=max(f[j],f[j-c[i]]+w[i]);
       }
       printf("%d\n",f[v]);
   }
   return 0;
}