用记忆化搜索的方法做01背包题

     今天看题的时候，看到了早就做过的01背包题，想想当时刚开始学的时候，用的是记忆化搜索，这种方法对时间复杂度上的要求比较高，相信很多大牛们都不用了吧，但我还是写了一下，算是一种加强记忆吧，以后有想看的也可以在这里找到：

一道hdu上的01背包裸题：http://acm.hdu.edu.cn/showproblem.php?pid=2602

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 41109    Accepted Submission(s): 17110

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 ?

 

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 2³¹).

 

Sample Input

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

 

下面是AC代码：更高效的01背包代码网上挺多的，以下就不给出了。

#include <iostream>#include <cstring>#include <cstdio>#define MAXN 1000+10using namespace std;int N,V;long long dp[1010][1010];//dp[i][j]取第i件物品时，背包容量为j时价值的最大值struct node{    int v;    int w;}a[1010];long long dfs(int i,int j){    long long ans;    if(i<=0){        if(j>=a[0].w)            ans=a[0].v;        else            ans=0;    }    else if(dp[i][j]!=-1)        ans=dp[i][j];    else{        if(j>=a[i].w)            ans=max(dfs(i-1,j),dfs(i-1,j-a[i].w)+a[i].v);        else            ans=dfs(i-1,j);    }    dp[i][j]=ans;    return dp[i][j];}int main(){    int T;    scanf("%d",&T);    while(T--)    {        memset(dp,-1,sizeof(dp));        scanf("%d%d",&N,&V);        for(int i=0;i<N;i++) scanf("%d",&a[i].v);        for(int i=0;i<N;i++) scanf("%d",&a[i].w);                printf("%I64d\n",dfs(N-1,V));    }    return 0;}