poj1742 单调队列优化多重背包

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如题：http://poj.org/problem?id=1742

Coins

Time Limit: 3000MS Memory Limit: 30000KTotal Submissions: 31258 Accepted: 10640

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

3 101 2 4 2 1 12 51 4 2 10 0

Sample Output

Source

LouTiancheng@POJ

由这一题的M,N太大，使用二进制拆分的O(MN∑logni)直接wa，于是想起以前见过的单调队列去均摊复杂度到O(MN)，就是将容量拆分成余数+k*v[i]，找出单调性，入队，每次在窗口大小为n,n=min(num[i],V/v[i])里取出最大值更新。又去查了资料。终于基本搞清楚了。

这篇文章讲得很好：http://blog.csdn.net/flyinghearts/article/details/5898183，大家点进去看吧，贴上我借鉴作者部分代码后的AC代码。

还有就是一点，这一题只能用f[i]表示当前价值i能不能装出来，而使用平常的单调队列去算价值==背包容量也会超。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define MAXV 100005
#define MAXN 105
#define max(a,b)(a>b?a:b)

bool a[MAXV];
int w[MAXN];
int num[MAXN];
bool f[MAXV];

void multipack(int V,int v,int n)
{
int i,j,k;;
if(n==1)
{
  for(i=V;i>=v;i--)
   f[i]=f[i]||f[i-v];
  return;
}
if(n*v>=V)
{
  for(i=v;i<=V;i++)
   f[i]=f[i]|f[i-v];
  return;
}
for(j=0;j<v;j++)
{
  int sum=0;
  bool *p1=a,*p2=a-1;
  for(k=j;k<=V;k+=v)
  {
   if(p2==p1+n)
    sum-=*p1++;
   *++p2=f[k];
   sum+=f[k];
   if(sum!=0)
    f[k]=1;
  }
}
}

int main()
{
// freopen("C:\\1.txt","r",stdin);
int n,m;
while(~scanf("%d%d",&n,&m))
{
  if(n==0&&m==0)
   break;
  int i;
  for(i=0;i<n;i++)
   scanf("%d",&w[i]);
  for(i=0;i<n;i++)
   scanf("%d",&num[i]);
  memset(f,false,sizeof(f));
  f[0]=true;
  for(i=0;i<n;i++)
   multipack(m,w[i],num[i]);
  int res=0;
  for(i=1;i<=m;i++)
   if(f[i]==true)
    res++;
  cout<<res<<endl;
}
return 0;
}

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