F

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 ? 

InputThe 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.OutputOne 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
一个很简单的背包问题，但是用dfs记忆化搜索也可做，并且复杂度相同

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=1E3+10;
int n,m;
ll v[N];
ll w[N];
ll dp[N][N];
ll dfs(int x,int y){
    ll ans=0; 
    if(x<=0) return 0;
    if(dp[x][y]) return dp[x][y];
    if(w[x]>y) ans=dfs(x+1,y);
    else {
        ans=max(dfs(x+1,y),dfs(x+1,y-w[x])+v[x]);
    }
    return dp[x][y]=ans;
} 
int main(){
    int t;
    cin>>t;
    while(t--){
        cin>>n>>m;
        for(int i=1;i<=n;i++){
            cin>>v[i];
        }
        for(int j=1;j<=n;j++)
            cin>>w[j];
        memset(dp,0,sizeof(dp));
        cout<<dfs(1,m)<<endl; 
    }
    
    return 0;
}

写dfs的时候一定要清楚它的返回值的意义。