Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 39532 Accepted Submission(s): 16385
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 ?
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.
題解:
裸的01揹包,經典例題。
參考代碼:
#include<stdio.h>
#define M 1111
#include<string.h>
int max(int a,int b)
{
return a>b?a:b;
}
int main()
{
int t,a[M],b[M],dp[M];
scanf("%d",&t);
while(t--)
{
int n,v;
memset(dp,0,sizeof(dp));
scanf("%d%d",&n,&v);
for(int i=0;i<n;i++)
scanf("%d",&b[i]);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
for(int i=0;i<n;i++)
{
for(int j=v;j>=a[i];j--)
{
dp[j]=max(dp[j],dp[j-a[i]]+b[i]);
}
}
printf("%d\n",dp[v]);
}
return 0;
}