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 ?
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.
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).
total value (this number will be less than 231).
Sample Input
1
5 10
1 2 3 4 5
5 4 3 2 1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14
#include<cstdio>
#include<algorithm>
using namespace std;
long long dp[+];
int w[],p[];
int main()
{
int t,N,W;
scanf("%d",&t);
while(t--)
{
scanf("%d %d",&N,&W);
for(int i=;i<=N;i++) //
scanf("%d",&p[i]);
for(int i=;i<=N;i++) //
scanf("%d",&w[i]);
for(int i=;i<=W;i++)
dp[i]=;
for(int i=;i<=N;i++) // 这几个地方i开头值注意一样 错了几次..
{
for(int j=W;j>=w[i];j--)
{
dp[j]=max(dp[j],dp[j-w[i]]+p[i]); //核心代码
}
}
printf("%lld\n",dp[W]); } return ;
}