HDU 2844-Coins(多重背包)

我就是我 2022-07-24 08:14 124阅读 0赞

# Coins #

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 11770 Accepted Submission(s): 4686**

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

Source

[2009 Multi-University Training Contest 3 - Host by WHU][]

Recommend

gaojie | We have carefully selected several similar problems for you: [1203][] [1171][] [1864][] [2845][] [2191][]

## 题目意思： ##

有一些硬币的面值和这些面值的硬币的个数，求这些硬币能组成不超过价值m的面值有多少种可能？

## 解题思路： ##

多重背包。  
可以分成两部分考虑：  
①如果面值和数量总值大于m，可以看成完全背包来处理；  
②否则，分解成01背包，增加一个k表示取的硬币数量。  
dp\[i\]数组表示取硬币，所组成的价值为i。

#include<iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    #define MAX 100000
    int dp[MAX];
    int c[MAX],w[MAX];
    int n,m;
    
    void ZeroOnePack(int cost,int wei)//01背包
    {
        int i;
        for(i = m; i>=cost; i--)
            dp[i]=max(dp[i],dp[i-cost]+wei);
    }
    
    void CompletePack(int cost,int wei)//完全背包
    {
        int i;
        for(i = cost; i<=m; i++)
            dp[i] = max(dp[i],dp[i-cost]+wei);
    }
    
    void MultiplePack(int cost,int wei,int cnt)//多重背包
    {
        if(m<=cnt*cost)//完全背包
        {
            CompletePack(cost,wei);
            return ;
        }
        else
        {
            int k;
            for(k=1; k<=cnt; ++k)
            {
                ZeroOnePack(k*cost,k*wei);//01背包
                cnt=cnt-k;
            }
            ZeroOnePack(cnt*cost,cnt*wei);//01背包
        }
    }
    
    int main()
    {
        while(~scanf("%d%d",&n,&m),n+m)
        {
            int i;
            for(i = 0; i<n; i++)
                scanf("%d",&c[i]);
            for(i = 0; i<n; i++)
                scanf("%d",&w[i]);
            memset(dp,0,sizeof(dp));
            for(i=0; i<n; i++)
                MultiplePack(c[i],c[i],w[i]);
            int ans=0;
            for(i=1; i<=m; i++)
                if(dp[i]==i)
                    ++ans;
            printf("%d\n",ans);
        }
        return 0;
    }

写在主函数里面就TLE了……就像这样↓

#include<iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    #define MAXN 105
    int a[MAXN],c[MAXN];
    int dp[MAXN];
    //dp[i]：价值为i的物品
    int main()
    {
        int n,m;
        while(~scanf("%d%d",&n,&m),n+m)
        {
            memset(dp,0,sizeof(dp));
            memset(a,0,sizeof(a));
            memset(c,0,sizeof(c));
            int i,j,k,ans=0;
            for(i=0; i<n; ++i)
                scanf("%d",&a[i]); //value
            for(i=0; i<n; ++i)
                scanf("%d",&c[i]); //number
            for(i=0; i<n; ++i)
            {
                if(a[i]*c[i]>=m)//完全背包
                    for(j=a[i]; j<=m; j++)
                        dp[j]=max(dp[j],dp[j-a[i]]+a[i]);
                else//01背包
                {
                    for(k=1; k<=c[i]; k<<=1) //数量
                    {
                        for(j=m; j>=k*a[i]; --j)//价值
                            dp[j]=max(dp[j],dp[j-k*a[i]]+k*a[i]);
                        c[i]-=k;//减去用过的数量
                    }
                    for(j=m; j>=a[i]*c[i]; --j)
                        dp[j]=max(dp[j],dp[j-a[i]*c[i]]+a[i]*c[i]);
                }
            }
            for(i=1; i<=m; ++i)
                if(dp[i]==i) ++ans;
            printf("%d\n",ans);
        }
        return 0;
    }

[2009 Multi-University Training Contest 3 - Host by WHU]: http://acm.hdu.edu.cn/search.php?field=problem&key=2009+Multi-University+Training+Contest+3+-+Host+by+WHU+&source=1&searchmode=source
[1203]: http://acm.hdu.edu.cn/showproblem.php?pid=1203
[1171]: http://acm.hdu.edu.cn/showproblem.php?pid=1171
[1864]: http://acm.hdu.edu.cn/showproblem.php?pid=1864
[2845]: http://acm.hdu.edu.cn/showproblem.php?pid=2845
[2191]: http://acm.hdu.edu.cn/showproblem.php?pid=2191