Bone Collector--hdu2602（01背包）

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 2 ³¹).

 

 

Sample Input

1

5 10

1 2 3 4 5

5 4 3 2 1

 

 

Sample Output

14

 

 

Author

Teddy

 

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

 

题意：给你n个骨头的价值和体积，求体积为V时最多有多少价值；

01背包问题:

递推公式：dp[i][j]=dp[i+1][j];(j<w[i]);

　　　　　dp[i][j]=max(dp[i+1][j],dp[i+1][j-w[i]]+v[i]);

代码如下：

 

关于i的逆向循环；

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<algorithm>

using namespace std;

#define N 1100

   int dp[N][N];

int main()

{

    int T,i,j,V,n,v[N],w[N];



    scanf("%d",&T);

    while(T--)

    {

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

        scanf("%d%d",&n,&V);

        for(i=0;i<n;i++)

                scanf("%d",&v[i]);

        for(i=0;i<n;i++)

                scanf("%d",&w[i]);

        for(i=n-1;i>=0;i--)

        {

            for(j=0;j<=V;j++)

            {

                if(j<w[i])

                    dp[i][j]=dp[i+1][j];

                else

                    dp[i][j]=max(dp[i+1][j],dp[i+1][j-w[i]]+v[i]);

            }

        }

        printf("%d\n",dp[0][V]);

    }

    return 0;

}

View Code

下面是关于i的正向循环；

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<algorithm>

using namespace std;

#define N 1100

   int dp[N][N];

int main()

{

    int T,i,j,V,n,v[N],w[N];



    scanf("%d",&T);

    while(T--)

    {

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

        scanf("%d%d",&n,&V);

        for(i=0;i<n;i++)

                scanf("%d",&v[i]);

        for(i=0;i<n;i++)

                scanf("%d",&w[i]);

        for(i=0;i<n;i++)

        {

            for(j=0;j<=V;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][V]);

    }

    return 0;

}

View Code

用一维数组；

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<algorithm>

using namespace std;

#define N 1100

int dp[N];

int main()

{

    int T,i,j,V,n,v[N],w[N];



    scanf("%d",&T);

    while(T--)

    {

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

        scanf("%d%d",&n,&V);

        for(i=0;i<n;i++)

                scanf("%d",&v[i]);

        for(i=0;i<n;i++)

                scanf("%d",&w[i]);

        for(i=0;i<n;i++)

        {

            for(j=V;j>=w[i];j--)

            {

                dp[j]=max(dp[j],dp[j-w[i]]+v[i]);

            }

        }

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

    }

    return 0;

}

View Code