poj-1742 Coins（多重背包优化）

题目链接：http://poj.org/problem?id=1742

Coins

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 33242		Accepted: 11283

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

题意；n种硬币，目标金钱是m。每种硬币给你数目和价值，问在1~m之前最多能组成多少种钱。

思路：楼教主“男人八题”之一，此题看似简单，数据却十分变态。错的基本都是TLE。 二进制多重背包过不了，单调队列优化的多重背包也过不了，加了混合背包单调队列才卡过，只能说自己太弱了。对单调队列的理解不够深。

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
#define M 100010
#define N 110
int dp[M];
int b[M];
int l,r;
int n,m;
int s[N],flag[M];
int main()
{
    int num,sum;
    while(~scanf("%d %d",&n,&m)&&(n+m))
    {
        for(int i=1; i<=n; i++)
            scanf("%d",&s[i]);
        for(int i=1; i<=m; i++)
            dp[i]=0;
        dp[0]=1;
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&num);
            int v=s[i];
            if(num==1)
            {
                for(int j=m; j>=v; j--)
                    if(!dp[j]&&dp[j-v])
                        dp[j]=1;
            }
            else if(num*v>=m)
            {
                 for(int j=v; j<=m; j++)
                    if(!dp[j]&&dp[j-v])
                        dp[j]=1;
            }
            else
            {
                for(int d=0; d<v; d++)
                {
                    sum=0,l=1;r=0;
                    for(int j=d;j<=m;j+=v)
                    {
                        if(r-l+1>num)
                        sum-=b[l++];
                        b[++r]=dp[j];
                        sum+=dp[j];
                        if(!dp[j]&&sum)
                            dp[j]=1;
                    }
                }
            }
        }
        int t=0;
        for(int i=1;i<=m;i++)
            if(dp[i])
            t++;
        printf("%d\n",t);
    }
    return 0;
}