(优化的有限背包) poj 1742 Coins

最新推荐文章于 2021-02-05 10:03:25 发布

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本文介绍了一道经典的计算机编程问题——硬币组合问题。该问题要求计算在给定不同面额硬币及数量的情况下，能组成的不超特定总额的有效价格数量。通过动态规划方法，实现了高效的求解算法。

Coins

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 16723		Accepted: 5807

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
/*
题目大意：略；
题目解答：用化成2^k的方法复杂度为O(V*Σlog n[i]),优化后复杂度为o(NV)；
ps：楼天成出的题目
Source Code

Problem: 1742  User: wawadimu 
Memory: 920K  Time: 1313MS 
Language: C++  Result: Accepted 

Source Code
*/ 
#include<iostream>
#include<string>
using namespace std;
int dp[100001];
int A[101];//物品重量
int C[101];//物品数目
int num[100001];//当前使用第i种物品的数目
int n,m;
int main()
{
    int i,j,sum;
    while(scanf("%d%d",&n,&m)!=EOF && !(n==0 && m==0))
    {
        sum=0;
        for(i=1;i<=n;i++)
            scanf("%d",A+i);
        for(i=1;i<=n;i++)
            scanf("%d",C+i);
        memset(dp,0,sizeof(dp));
        dp[0]=1;//背包为空
        for(i=1;i<=n;i++)
        {
            memset(num,0,sizeof(num));
            for(j=A[i];j<=m;j++)//多重背包正序计算
            {
                if(!dp[j] && dp[j-A[i]] && num[j-A[i]] < C[i])//可添加第i种物品，并且没有超出数目限制
                {
                    dp[j]=1;
                    sum++;
                    num[j]=num[j-A[i]]+1;//数目增加1
                }
            }
        }
        printf("%d/n",sum);

    }
    return 0;
}