hdu2602-01背包

Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 48440 Accepted Submission(s): 20204

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

Author

Teddy

Source

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

简单的01背包

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int vol[1010],val[1010],dp[1010];
int main(){
int t,n,v,i,j;
//freopen("in.txt","r",stdin);
while(scanf("%d",&t)!=EOF){
while(t--){
memset(dp,0,sizeof(dp));
scanf("%d%d",&n,&v);
for ( i = 0; i < n; ++i)
{
scanf("%d",&val[i]);
}
for ( i = 0; i < n; ++i)
{
scanf("%d",&vol[i]);
}

for ( i = 0; i < n; ++i)
{
for ( j = v; j >=vol[i]; j--)
{
dp[j]=max(dp[j],dp[j-vol[i]]+val[i]);
}
}
printf("%d\n", dp[v]);
}
}

return 0;
}