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.
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
Recommend
lcy
思路：01背包问题
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
int main()
{
int n,v,t,i,j,value[1005],volume[1005],dp[1005];
scanf("%d",&t);
while(t--)
{
memset(value,0,sizeof(value));
memset(volume,0,sizeof(volume));
memset(dp,0,sizeof(dp));
scanf("%d%d",&n,&v);
for(i=1;i<=n;i++)
scanf("%d",&value[i]);
for(i=1;i<=n;i++)
scanf("%d",&volume[i]);
for(i=1;i<=n;i++)
for(j=v;j>=volume[i];j--)
if(dp[j]<dp[j-volume[i]]+value[i])
dp[j]=dp[j-volume[i]]+value[i];
printf("%d\n",dp[v]);
}
return 0;
}