HDU 2602 Bone Collector (01背包问题)

原题代号：HDU 2602

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=2602

原题描述：

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

Sample Output

14

题目大意：骨头收集者在收集骨头，有容量为v的袋子，n个骨头，给出每个骨头的价值和体积。问最大能收集的价值是多少。（01背包问题）

解法一：时间，空间复杂度均为O(n²)

# include <stdio.h>
# include <string.h>
# include <stdlib.h>
# include <iostream>
# include <fstream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <math.h>
# include <algorithm>
using namespace std;
# define pi acos(-1.0)
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define For(i,n,a) for(int i=n; i>=a; --i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define Fo(i,n,a) for(int i=n; i>a ;--i)
typedef long long LL;
typedef unsigned long long ULL;

int a[+],b[+],dp[+][+];

int main()
{
    int t,n,v;
    cin>>t;
    while(t--)
    {
        cin>>n>>v;
        FOR(i,,n)cin>>b[i];
        FOR(i,,n)cin>>a[i];
        mem(dp,);
        FOR(i,,n)FOR(j,,v)
        {
            if(a[i]<=j)dp[i][j]=max(dp[i-][j],dp[i-][j-a[i]]+b[i]);
            else dp[i][j]=dp[i-][j];
        }
        cout<<dp[n][v]<<endl;
    }
    return ;
}

解法二：时间复杂度O(n²)，空间复杂度O(n)

# include <stdio.h>
# include <string.h>
# include <stdlib.h>
# include <iostream>
# include <fstream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <math.h>
# include <algorithm>
using namespace std;
# define pi acos(-1.0)
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define For(i,n,a) for(int i=n; i>=a; --i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define Fo(i,n,a) for(int i=n; i>a ;--i)
typedef long long LL;
typedef unsigned long long ULL;

int a[+],b[+],dp[+];

int main()
{
    int t,n,v;
    cin>>t;
    while(t--)
    {
        int ans=;
        cin>>n>>v;
        FOR(i,,n)cin>>b[i];
        FOR(i,,n)cin>>a[i];
        mem(dp,);
        FOR(i,,n)For(j,v,)
        {
            if(a[i]<=j)dp[j]=max(dp[j],dp[j-a[i]]+b[i]);
        }
        cout<<dp[v]<<endl;
    }
    return ;
}