HDOJ 2602

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14134    Accepted Submission(s): 5585

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

//下面的不对 
#include <iostream>
#include <cstring>
using namespace std;
typedef struct Node
{
    int w,value;
}Node;
Node bag[1010];
int f[1010][1010];
int main()
{
    int i,j,k,t,T;
    cin>>T;
    while(T--)
    {
        int n,c;
        cin>>n>>c;
        memset(bag,0,sizeof(bag));
        memset(f,0,sizeof(f));
        for(i=1;i<=n;i++)
            cin>>bag[i].value;
        for(i=1;i<=n;i++)
            cin>>bag[i].w;
        f[1][c] = bag[1].value;
        for(i=2;i<=n;i++)
        {
            if(c>=bag[i].w)
            {
                int temp = f[i-1][c-bag[i].w] + bag[i].value;
                f[i][c] = max(temp,f[i-1][c]);
            }
            else 
                f[i][c] = f[i-1][c];
        }
        cout<<f[n][c]<<endl;
    }
    return 0;
}
        

#include <iostream>
#include <cstring>
using namespace std;
int w[1010];
int value[1010];
int f[1010][1010];
int main()
{
    int i,j,k,t,T;
    cin>>T;
    while(T--)
    {
        int n,c;
        cin>>n>>c;
        memset(w,0,sizeof(w));
        memset(f,0,sizeof(f));
        memset(value,0,sizeof(value));
        for(i=1;i<=n;i++)
            cin>>value[i];
        for(i=1;i<=n;i++)
            cin>>w[i];
        for(i=1;i<=n;i++)
            for(j=0;j<=c;j++)
            if(j>=w[i])
            {
                f[i][j] = max(f[i-1][j],f[i-1][j-w[i]] + value[i]);
            }
            else
                f[i][j] = f[i-1][j];
                /*
                //原来没加这一条语句
                1
                5 0
                2 4 1 5 1
                0 0 1 0 0
                答案应该是12
                却输出6 
                */ 
            cout<<f[n][c]<<endl;
    }
    return 0;
}
        

//此时g++提交ac，c++wa，因为第二十九行的符号 
#include <iostream>
#include <cstring>
using namespace std;
int w[1010];
int value[1010];
int f[1010][1010];
int main()
{
    int i,j,k,t,T;
    cin>>T;
    while(T--)
    {
        int n,c;
        cin>>n>>c;
        memset(w,0,sizeof(w));
        memset(f,0,sizeof(f));
        memset(value,0,sizeof(value));
        for(i=1;i<=n;i++)
            cin>>value[i];
        for(i=1;i<=n;i++)
            cin>>w[i];
        for(i=1;i<=n;i++)
            for(j=0;j<=c;j++)
            {
                f[i][j] = f[i-1][j];
                if(j>=w[i])
                {
                    f[i][j] >?= f[i-1][j-w[i]] + value[i];
                }
            }
            cout<<f[n][c]<<endl;
    }
    return 0;
}
        

//滚动数组优化空间和时间复杂度 ,时间上不太明显 
#include <iostream>
#include <cstring>
using namespace std;
int w[1010];
int value[1010];
int f[1010];
int main()
{
    int i,j,k,t,T;
    cin>>T;
    while(T--)
    {
        int n,c;
        cin>>n>>c;
        memset(w,0,sizeof(w));
        memset(f,0,sizeof(f));
        memset(value,0,sizeof(value));
        for(i=1;i<=n;i++)
            cin>>value[i];
        for(i=1;i<=n;i++)
            cin>>w[i];
        for(i=1;i<=n;i++)
            for(j=c;j>=w[i];j--)
            {
                f[j] >?= f[j-w[i]] + value[i];
            }
            cout<<f[c]<<endl;
    }
    return 0;
} 
//注意一条：不可用结构体存储重量和价值