POJ-1742 Coins
题面
Problem 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.
题意
钱有多少组合方式，组成的价格不同。
上来，诶 多重背包 二进制优化 美滋滋。
TLE。。MMP。。。
一查，居然是啥真男人八题。。。
用单调队列来优化多重背包。论文请查看2009年的徐持衡的论文《浅谈几类背包题》
太晚了先睡了，POJ比openjudge慢好多啊QWQ
代码
#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
#include <deque>
#include <cstdio>
using namespace std;
int n,m;
int dp[100010];
pair<int,int> p[110];
int ans;
deque<pair<int,int> > q;
int main()
{
while (scanf("%d%d",&n,&m)==2)
{
if (n==0 && m==0) return 0;
memset(dp,0,sizeof dp);
dp[0]=1;
for (int i=1;i<=n;i++) scanf("%d",&p[i].first);
for (int i=1;i<=n;i++) scanf("%d",&p[i].second);
for (int i=1;i<=n;i++)
{
if (p[i].second==1)
{
for (int j=m-p[i].first;j>=0;j--) if (dp[j]) dp[j+p[i].first]=1;
}
else if (p[i].first*p[i].second>=m)
{
for (int j=0;j<=m;j++) if (dp[j] && j+p[i].first<=m) dp[j+p[i].first]=1;
}
else
{
for (int j=0;j<p[i].first;j++)
{
while (!q.empty()) q.pop_front();
for (int k=j;k<=m;k+=p[i].first)
{
if (!q.empty()) {
if (k-q.front().first>p[i].first*p[i].second) q.pop_front();
}
if (dp[k])
{
q.push_back(make_pair(k,dp[k]));
}
else if (!q.empty()) dp[k]=1;
}
}
}
}
ans=0;
for (int i=1;i<=m;i++) ans+=dp[i];
printf("%d
",ans);
}
}