Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 44257    Accepted Submission(s): 18442

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

1

5 10

1 2 3 4 5

5 4 3 2 1

01背包模板题

/*
    Accepted    2602    78MS    5360K    552 B    G++
*/
#include"cstdio"
#include"cstring"
#include"algorithm"
using namespace std;
const int MAXN=1005;
int dp[MAXN][MAXN];
int n,W;
int v[MAXN],w[MAXN];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&W);
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++)    scanf("%d",&v[i]);
        for(int i=0;i<n;i++)    scanf("%d",&w[i]);

        for(int i=0;i<n;i++)
            for(int j=0;j<=W;j++)
                if(j<w[i])    dp[i+1][j]=dp[i][j];
                else dp[i+1][j]=max(dp[i][j],dp[i][j-w[i]]+v[i]);

        printf("%d\n",dp[n][W]);
    }

    return 0;
}

贴一个记忆化搜索

/*
    Accepted    2602    327MS    5432K    613 B    G++
*/
#include"cstdio"
#include"cstring"
#include"algorithm"
using namespace std;
const int MAXN=1005;
int dp[MAXN][MAXN];
int n,W;
int v[MAXN],w[MAXN];
int dfs(int i,int wei)
{
    if(dp[i][wei])    return dp[i][wei];
    if(i==n)    return 0;
    int res;
    if(wei>=w[i])    res=max(dfs(i+1,wei-w[i])+v[i],dfs(i+1,wei));
    else    res=dfs(i+1,wei);
    return dp[i][wei]=res;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&W);
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++)    scanf("%d",&v[i]);
        for(int i=0;i<n;i++)    scanf("%d",&w[i]);
        printf("%d\n",dfs(0,W));
    }

    return 0;
}