Hdu 1864 Maximum Reimbursement

maximum reimbursement

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 25076    Accepted Submission(s): 7716

Problem Description

Existing funds can be reimbursed for a certain amount of invoices. The types of invoices that are allowed to be reimbursed include buying books ( Type A), stationery (Type B), and business trips (Type C). . Now please write a program to find out the maximum reimbursement amount that can be reimbursed within a given amount of invoices.

 

 

Input

Test input contains several test cases. Line 1 of each test case contains two positive numbers Q and N, where Q is the given reimbursement amount and N (<=30) is the number of invoices. Followed by N lines of input, the format of each line is:
m Type_1:price_1 Type_2:price_2 ... Type_m:price_m
where the positive integer m is the number of items issued on this invoice, and Type_i and price_i are the types of the item i and value. The item type is represented by a capital letter. When N is 0, all input ends, and the corresponding result is not output.

 

 

Output

Output 1 line for each test case , the maximum amount that can be reimbursed, accurate to 2 decimal places.

 

 

Sample Input

200.00 3

2 A:23.50 B:100.00

1 C:650.00

3 A:59.99 A:120.00 X:10.00

1200.00 2

2 B:600.00 A:400.00

1 C:200.50

1200.50 3

2 B:600.00 A:400.00

1 C:200.50

1 A:100.00

100.00 0

 

 

Sample Output

123.50

1000.00

1200.50

#include <stdio.h>  
#include <algorithm>  
#include <string.h>  
using namespace std;  
  
int dp[3000050];
  
intmain()  
{  
    char ch;  
    double x,y;  
    int sum,a,b,c,money[35],v;  
    int t,i,j,k;  
    while(~scanf("%lf%d",&x,&t),t)  
    {  
        sum = (int)(x*100);//Convert decimals to integers memset(money,0,sizeof(money));  
        memset(dp,0,sizeof(dp));  
        int l = 0;  
        for(i = 0; i<t; i++)  
        {  
            scanf("%d",&k);  
            a = b = c = 0;  
            int flag = 1;  
            while(k--)  
            {  
                scanf(" %c:%lf",&ch,&y);  
                v = (int)(y*100);  
                if(ch == 'A' && a+v<=60000)  
                    a+=v;  
                else if(ch == 'B' && b+v<=60000)  
                    b+=v;  
                else if(ch == 'C' && c+v<=60000)  
                    c+=v;  
                else  
                    flag = 0;  
            }  
            if(a+b+c<=100000 && a<=60000 && b<=60000 && c<=60000 && flag)//According to the meaning of the title, these conditions must be met money[l++] = a+b+c ;  
        }  
        for(i = 0; i<=l; i++)  
        {  
            for(j = sum; j>=money[i]; j--)  
                    dp[j] = max(dp[j],dp[j-money[i]]+money[i]);  
        }  
        printf("%.2lf\n",dp[sum]/100.0);  
    }  
  
    return 0;  
}