POJ1276 Cash Machine(动态规划)

Cash Machine

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 28982 Accepted: 10375

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.

Sample Input

735 3  4 125  6 5  3 350633 4  500 30  6 100  1 5  0 1735 00 3  10 100  10 50  10 10

Sample Output

73563000

Hint

The first data set designates a transaction where the amount of cash requested is @735. The machine contains 3 bill denominations: 4 bills of @125, 6 bills of @5, and 3 bills of @350. The machine can deliver the exact amount of requested cash.

In the second case the bill supply of the machine does not fit the exact amount of cash requested. The maximum cash that can be delivered is @630. Notice that there can be several possibilities to combine the bills in the machine for matching the delivered cash.

In the third case the machine is empty and no cash is delivered. In the fourth case the amount of cash requested is @0 and, therefore, the machine delivers no cash.

思路:

由于DP训练得过少，所以一开始没有什么思路，看了某前辈的思路，才勉勉强强写出来，看来还要多练，基本思路就是多重背包的运用，多重背包就是每个物品的件数是有限制的，然后给定背包能承受的重量，求能拿到的最大价值。本题可以定义一个Max来更新最大值，Max的范围一定小于给定的值，比如输入735那么一定小于735，每次更新Max的值，运用了动态规划的思想。

代码:

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int maxn = 1001;const int maxn1 = 100010;int c[maxn];int v[maxn];int dp[maxn1];int Max;void DP(int m,int total){    int i,j,k;    for(i=1; i<=m; i++)    {        for(k=Max; k>=0; k--)        {            if(dp[k])  //如果没有访问过，访问过就直接返回            {                for(j=0; j<=c[i]; j++)                {                    int temp = j * v[i] + k;  //当前能达到的值                    if(temp > total)  //如果大于目标值那么不满足条件，继续找                        continue;                    dp[temp] = 1; //访问过，记录                    if(temp > Max) //如果当前能达到的值大于Max更新                    {                        Max = temp;                    }                }            }        }    }}int main(){    int total;    int i;    int m;    //freopen("111","r",stdin);    while(cin>>total>>m)    {        Max = 0;        if(m == 0)        {            cout<<"0"<<endl;            continue;        }        for(i=1; i<=m; i++) //一共m种货币        {            cin>>c[i]>>v[i];        }        memset(dp,0,sizeof(dp));        dp[0] = 1;  //才能够进入判定条件        DP(m,total);        cout<<Max<<endl;    }    return 0;}