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 2 31).
Sample Input
1 5 10 1 2 3 4 5 5 4 3 2 1
Sample Output
14
分析:01背包问题,动态方程:dp[j] = max(dp[j],dp[j-b[i]]+a[i])
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 1005
int a[maxn],b[maxn];
int dp[maxn];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,m;
scanf("%d %d",&n,&m);
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(dp,0,sizeof(dp));
for(int i = 1; i <= n; i ++)
scanf("%d",&a[i]);
for(int i = 1; i <= n; i ++)
scanf("%d",&b[i]);
for(int i = 1; i <= n; i ++)
for(int j = m; j >= b[i]; j--)
{
dp[j] = max(dp[j],dp[j-b[i]]+a[i]);
}
printf("%d\n",dp[m]);
}
}