POJ1276 Cash Machine 【解法二】

Description A Bank plans to install a machine for cash withdrawal. The
machine is able to deliver appropriate @ bills for a requested cash
amount. The machine uses exactly N distinct bill denominations, say
Dk, k=1,N, and for each denomination Dk the machine has a supply of nk
bills. For example,

N=3, n1=10, D1=100, n2=4, D2=50, n3=5, D3=10

means the machine has a supply of 10 bills of @100 each, 4 bills of
@50 each, and 5 bills of @10 each.

Call cash the requested amount of cash the machine should deliver and
write a program that computes the maximum amount of cash less than or
equal to cash that can be effectively delivered according to the
available bill supply of the machine.

Notes: @ is the symbol of the currency delivered by the machine. For
instance, @ may stand for dollar, euro, pound etc.

Input The program input is from standard input. Each data set in the
input stands for a particular transaction and has the format:

cash N n1 D1 n2 D2 … nN DN

where 0 <= cash <= 100000 is the amount of cash requested, 0 <=N <= 10
is the number of bill denominations and 0 <= nk <= 1000 is the number
of available bills for the Dk denomination, 1 <= Dk <= 1000, k=1,N.
White spaces can occur freely between the numbers in the input. The
input data are correct.

Output For each set of data the program prints the result to the
standard output on a separate line as shown in the examples below.

解法一见【这里】。
单调队列优化多重背包。
但是由于不是求最优解，只是判断可行性，所以实际上不需要建队列，只保存最近的可行解即可。

#include<cstdio>#include<cstring>const int oo=0x3f3f3f3f;bool dp[15][100010];int main(){    int i,j,k,m,n,p,q,x,y,z,v,num,hd,tl;    while (scanf("%d%d",&m,&n)==2)    {        memset(dp,0,sizeof(dp));        dp[0][0]=1;        for (i=1;i<=n;i++)        {            scanf("%d%d",&num,&v);            for (j=0;j<v;j++)            {                p=-oo;                for (k=0;k*v+j<=m;k++)                {                    if (k-p<=num) dp[i][k*v+j]=dp[i-1][p*v+j];                    dp[i][k*v+j]|=dp[i-1][k*v+j];                    if (dp[i-1][k*v+j]) p=k;                }            }        }        for (i=m;;i--)          if (dp[n][i])          {            printf("%d\n",i);            break;          }    }}