POJ 1472 Coins （多重背包+滾動數組）

Coins Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 25827   Accepted: 8741 Description People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change) and he known the price would not more than m.But he didn't know the exact price of the watch. You are to write a program which reads n,m,A1,A2,A3...An and C1,C2,C3...Cn corresponding to the number of Tony's coins of value A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay use these coins. Input The input contains several test cases. The first line of each test case contains two integers n(1<=n<=100),m(m<=100000).The second line contains 2n integers, denoting A1,A2,A3...An,C1,C2,C3...Cn (1<=Ai<=100000,1<=Ci<=1000). The last test case is followed by two zeros. Output For each test case output the answer on a single line. Sample Input 3 10 1 2 4 2 1 1 2 5 1 4 2 1 0 0 Sample Output 8 4   題意：給出幾種面值的錢幣和對應的個數，看能否湊出1-m中的各個面值。   方法一：多重背包

import java.io.*;  
import java.util.*;  
/* 
 * 
 * author : deng_hui_long  
 * Date   : 2013-8-31 
 * 
*/  
public class Main {  
    int n,m;  
    int dp[]=new int[1000001];  
    public static void main(String[] args) {  
        new Main().work();  
    }  
    void work(){  
        Scanner sc=new Scanner(new BufferedInputStream(System.in));  
        while(sc.hasNext()){  
            n=sc.nextInt();  
            m=sc.nextInt();  
            if(n==0&&m==0)  
                break;  
            Node node=new Node();  
            Arrays.fill(dp, 0);  
              
            for(int i=0;i<n;i++){  
                node.a[i]=sc.nextInt();  
            }  
              
            for(int i=0;i<n;i++){  
                node.c[i]=sc.nextInt();  
            }  
              
            for(int i=0;i<n;i++){  
                multiplePack(node.a[i],node.a[i],node.c[i]);  
            }  
            int ans=0;  
            for(int i=1;i<=m;i++){  
                if(dp[i]==i)  
                    ans++;  
            }  
            System.out.println(ans);  
        }  
    }  
    //多重背包  
    void multiplePack(int cost,int weight,int amount){  
        if(cost*amount>=m)//大於最大價值，按完全背包處理  
            completePack(cost,weight);  
        else{//小於最大價值，按01背包處理  
            int k=1;  
            while(k<amount){  
                zeroOnePack(k*cost,k*weight);  
                amount-=k;  
                k<<=1;//右一位，表示乘以2  
            }  
            zeroOnePack(amount*cost,amount*weight);  
        }  
    }  
    //完全背包  
    void completePack(int cost,int weight){  
        for(int i=cost;i<=m;i++){  
            dp[i]=Math.max(dp[i],dp[i-cost]+weight);  
        }  
    }  
    //01背包  
    void zeroOnePack(int cost,int weight){  
        for(int i=m;i>=cost;i--){  
            dp[i]=Math.max(dp[i],dp[i-cost]+weight);  
        }  
    }  
    class Node{  
        int a[]=new int[n];  
        int c[]=new int[n];  
    }  
}  
 

方法二：滾動數組

import java.io.*;  
import java.util.*;  
/* 
 * 
 * author : deng_hui_long  
 * Date   : 2013-8-31 
 * 
*/  
public class Main {  
    int n,m,MAX=100001;  
    boolean dp[]=new boolean[MAX];  
    public static void main(String[] args) {  
        new Main().work();  
    }  
    void work(){  
        Scanner sc=new Scanner(new BufferedInputStream(System.in));  
        while(sc.hasNext()){  
            n=sc.nextInt();  
            m=sc.nextInt();  
            if(n==0&&m==0)  
                break;  
            Node node=new Node();  
            for(int i=0;i<n;i++){  
                node.a[i]=sc.nextInt();  
            }  
            for(int i=0;i<n;i++){  
                node.c[i]=sc.nextInt();  
            }  
            Arrays.fill(dp,false);  
            dp[0]=true;  
            int ans=0;  
            //滾動數組  
            for(int i=0;i<n;i++){  
                int u[]=new int[MAX];  
                for(int j=node.a[i];j<=m;j++){  
                    if(!dp[j]&&dp[j-node.a[i]]&&node.c[i]>u[j-node.a[i]]){  
                        dp[j]=true;  
                        u[j]=u[j-node.a[i]]+1;  
                        ans++;  
                    }  
                }  
            }  
            System.out.println(ans);  
        }  
    }  
    class Node{  
        int a[]=new int[n];  
        int c[]=new int[n];  
    }  
}