hdu 2602 Bone Collector 01背包

Bone Collector

Time Limit: 1 Sec  Memory Limit: 256 MB

题目连接

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

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 2 ³¹).

Sample Input

1
5 10
1 2 3 4 5
5 4 3 2 1

 

Sample Output

14

 

HINT

 

题意

01背包裸题

题解：

01背包，滚动数组优化一下

代码：

 

//qscqesze

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

#include <map>

#include <stack>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define maxn 200001

#define mod 10007

#define eps 1e-9

int Num;

char CH[20];

//const int inf=0x7fffffff;   //нчоч╢С

const int inf=0x3f3f3f3f;

/*



inline void P(int x)

{

    Num=0;if(!x){putchar('0');puts("");return;}

    while(x>0)CH[++Num]=x%10,x/=10;

    while(Num)putchar(CH[Num--]+48);

    puts("");

}

*/

//**************************************************************************************

inline ll read()

{

    int x=0,f=1;char ch=getchar();

    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

    return x*f;

}

inline void P(int x)

{

    Num=0;if(!x){putchar('0');puts("");return;}

    while(x>0)CH[++Num]=x%10,x/=10;

    while(Num)putchar(CH[Num--]+48);

    puts("");

}



struct node

{

    int v,vl;

};

node a[maxn];

int dp[maxn];

int main()

{

    int t=read();

    for(int cas=1;cas<=t;cas++)

    {

        memset(a,0,sizeof(a));

        memset(dp,0,sizeof(dp));

        int n=read(),v=read();

        for(int i=1;i<=n;i++)

            a[i].v=read();

        for(int i=1;i<=n;i++)

            a[i].vl=read();

        for(int i=1;i<=n;i++)

        {

            for(int j=v;j>=a[i].vl;j--)

            {

                dp[j]=max(dp[j],dp[j-a[i].vl]+a[i].v);

            }

        }

        cout<<dp[v]<<endl;



    }

}