HDU 2844 Coins (多重背包)

Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13004    Accepted Submission(s): 5208

Problem Description

Whuacmers use coins.They have coins of value A1,A2,A3...An Silverland dollar. One day Hibix opened purse 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

 

题目大意：你有n种硬币，让你把这些硬币换成[1,m]，有哪些可以做到.

比如 

2 5

1 4 2 1

这两种硬币可以换成 1,2,4,5,6 但是6大于5，所以不符合。因此有4种符合。

思路：这题思路很简单，就是分别把为 i 属于 1 - m 当做容量为i的背包，用多重背包做就好了，因为背包问题所有的策略都是当前容量最多能装多少，最多就是装满，也就是如果能装满，那这个策略肯定可以装满。。。有一点不明白 为什么网上的代码都要分成多重背包跟完全背包呢。。。完全没必要啊，因为第二个for已经限制了for(int j = sum; j >= v[i]; j--)这样如果比背包容量大，直接跳过去了，不必担心数组下表<0

网上普遍的多重+完全

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>

using namespace std ;

int a[110],c[110] ;
int dp[100010] ;
bool vis[100010] ;
int n, m;

void zeroOnepack(int cost,int weight)
{
    for(int i = m ; i >= cost ; i--)
    dp[i] = max(dp[i],dp[i-cost]+weight) ;
}
void completepack(int cost ,int weight)
{
    for(int i = cost ; i <= m ; i++)
    dp[i] = max(dp[i],dp[i-cost]+weight) ;
}
void multiplepack(int cost,int weight,int amount)
{
    if(cost * amount >= m)
    {
        completepack(cost,weight) ;
        return  ;
    }
    else
    {
        int k = 1 ;
        while(k < amount)
        {
            zeroOnepack(k*cost,k*weight) ;
            amount -= k ;
            k = k*2 ;
        }
        zeroOnepack(amount*cost,amount*weight) ;
    }
}

int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        if(n == 0 && m==0) break ;
        for(int i = 0 ; i < n ; i++)
        scanf("%d",&a[i]) ;
        for(int j = 0 ; j < n ; j++)
        scanf("%d",&c[j]) ;
        memset(dp,0,sizeof(dp)) ;
        int cnt = 0 ;
        for(int i = 0 ; i < n ; i++)
        {
            multiplepack(a[i],a[i],c[i]) ;
        }
        for(int i = 1 ; i <= m ; i++)
        if(dp[i] == i)cnt++ ;
        printf("%d\n",cnt) ;
    }
    return 0 ;
}

直接多重转成二进制01就行

#include <iostream>  
#include <algorithm>  
#include <queue>  
#include <math.h>  
#include <stdio.h>  
#include <string.h>  
using namespace std;  
int M[100010];  
int main()  
{  
    int A[110],C[110];  
    int n,m;  
    int i,j,k;  
    while(scanf("%d %d",&n,&m),n|m){  
       for(i=1;i<=n;i++)  
          scanf("%d",&A[i]);  
       for(i=1;i<=n;i++)  
          scanf("%d",&C[i]);  
         
       memset(M,0,sizeof(M));  
         
       int temp;  
       for(i=1;i<=n;i++)  
       {  
            k=1;  
          while(k<=C[i])  
          {  
            C[i]-=k;  
            temp=k*A[i];  
            for(j=m;j>=temp;j--)  
                M[j]=max(M[j],M[j-temp]+temp);  
            k=k<<1;  
          }  
          if(C[i]>0)   
          {  
            k=C[i];  
            temp=k*A[i];  
            for(j=m;j>=temp;j--)  
                 M[j]=max(M[j],M[j-temp]+temp);  
                    
          }  
       }  
        
      k=0;  
     for(i=1;i<=m;i++)  
        if(M[i]==i)  
           k++;  
      printf("%d\n",k);  
    }  
    return 0;  
}