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): 32408 Accepted Submission(s): 13329
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
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int v[],w[];
int n,c;
int f[][];//对于dp(i,j)就表示可选物品为i…n背包容量为j(总重量)时背包中所放物品的最大价值。
void backpack()
{
int i,j;
for(i=;i<=n;i++)
for(j=;j<=c;j++)
{
if(i==)
f[i][j]=;
else
f[i][j]=f[i-][j];
if(j>=v[i])
f[i][j]=max(f[i][j],f[i-][j-v[i]]+w[i]);
}
printf("%d\n",f[n][c]);
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int i;
scanf("%d%d",&n,&c);
for(i=;i<=n;i++)
scanf("%d",&w[i]);
for(i=;i<=n;i++)
scanf("%d",&v[i]);
backpack();
}
return ;
}