poj 1742 多重背包（单调队列优化）

最新推荐文章于 2020-12-22 11:59:02 发布

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本文介绍了一种解决硬币组合支付问题的方法，该问题要求计算出在给定不同面值硬币及其数量的情况下，能够恰好支付的价格数目。通过使用动态规划技术，程序能够有效地找出所有可能的支付方案。

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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

Source

LouTiancheng@POJ

K爷的模板

 struct Queue//V-背包最大容量  c[i]第i种物品数量  v[i]第i种物品价值  w[i]第i种物品重量  注意定义全局变量  
{  
    int num,value;  
}que[250005];  
int head,tail;  
void enqueue (int x,int y)  
{  
    while (head<=tail && que[tail].value<y) tail--;  
    que[++tail].num=x;que[tail].value=y;  
}  
void multipack()  
{  
    int i,j,d;  
    memset(dp,0,sizeof(dp));  
    for (i=1 ; i<=n ; ++i)  
    {  
        if (c[i] > V/w[i]) c[i]=V/w[i];  
        for (d=0 ; d<w[i] ; ++d)  
        {  
            head=1;tail=0;  
            for (j=0 ; j<=(V-d)/w[i] ; ++j)  
            {  
                enqueue(j , dp[j*w[i]+d]-j*v[i]);  
                while (que[head].num<j-c[i] && head<=tail) head++;  
                dp[j*w[i]+d]=que[head].value+j*v[i];  
            }  
        }  
    }  
}  

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

const int maxV=100005;
const int maxn=105;
int c[maxn],w[maxn];
int n,V;
bool dp[maxV];

void ZeroOnePack(int cost){
for(int i=V;i>=cost;i--)dp[i]|=dp[i-cost];
}

void CompletePack(int cost){
for(int i=cost;i<=V;i++)dp[i]|=dp[i-cost];
}

void MultiplePack()
{
for(int i=1;i<=n;i++)
{

if(w[i]*c[i]>=V)CompletePack(w[i]);
else {
int k=1;
while(k<c[i]){
ZeroOnePack(k*w[i]);
c[i]-=k;
k<<=1;
}
ZeroOnePack(c[i]*w[i]);
}
}

}

int main(){

while(scanf("%d%d",&n,&V)&&n&&V)
{
for(int i=1;i<=n;i++)
cin>>w[i];
for(int i=1;i<=n;i++)
cin>>c[i];
for(int i=0;i<=V;i++)
dp[i]=0;
dp[0]=1;
MultiplePack();

int ans=0;
for(int i=1;i<=V;i++)if(dp[i])ans++;
printf("%d\n",ans);
}
return 0;

}