0-1リュックのまとめ

HDU 2602       Bone Collector
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 30300    Accepted Submission(s): 12477
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
 
 題意:t組のテストデータは、各組にn,mを入力し、物品の種類とリュックサックの大きさを表す.次の2行は、1行あたりn個の数で、最初の行は価値を表します.2行目は費用を表します.1つの品物は1回しか使えません!
      リュックサックが入る最大の価値を出力!
   
     コード(一):
     

// 0-1     



#include <stdio.h>

#include <string.h>

int w[1002], c[1002];

int dp[1002][1002];



int max(int a, int b)

{

    return a>b?a:b;

}



int main()

{

    int t;

    int i, j;

    scanf("%d", &t);

    int n, m;

    while(t--)

    {

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

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

        {

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

        }

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

        {

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

        }

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

            dp[i][0]=0;

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

            dp[0][j]=0;

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

        {

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

            {

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

                if( j>=c[i] )

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

            }

        }

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

    }

    return 0;

}

    コード(二)(劉汝佳の本に書いてある):
 

// 0-1     



#include <stdio.h>

#include <string.h>

int w[1002], c[1002];

int dp[1002][1002];



int max(int a, int b)

{

    return a>b?a:b;

}



int main()

{

    int t;

    int i, j;

    scanf("%d", &t);

    int n, m;

    while(t--)

    {

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

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

        {

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

        }

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

        {

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

        }

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

            dp[i][0]=0;

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

            dp[0][j]=0;

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

        {

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

            {

                dp[i][j] = (i==1?0:dp[i-1][j] );

                if( j>=c[i] )

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

            }

        }

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

    }

    return 0;

}