(优化的有限背包) poj 1742 Coins（转载）

 

最近在学习动态规划的背包问题，看了网上的背包九讲，感觉理解的比以前深入了一下，做了几道题，可是还是不够熟练，先练习一段时间，等熟练了再写个总结。

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; }