HDU -1864最大报销额(01背包)

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 
 6 int dp[3000050];//发票张数30*每张总额1000*放大倍数100 = 3000000
 7 int main()
 8 {
 9     char ch;
10     double x, y;
11     int sum, a, b, c, money[35], t;
12     int n, k;
13     while (~scanf("%lf %d", &x, &n) && n)
14     {
15         sum = (int)(x * 100);//将背包的容量放大为整数(这里就是报销的最大值)
16         memset(dp, 0, sizeof(dp));
17         memset(money, 0, sizeof(money));
18         int len = 0;//可报销的列表长度，也就是最后的n
19         for (int i = 0; i < n; i++)
20         {
21             scanf("%d", &k);
22             a = b = c = 0;//分别来保存a类型,b类型,c类型
23             bool flag = true;
24             while (k--)
25             {
26                 scanf(" %c:%lf", &ch, &y);
27                 t = (int)(y * 100);
28                 if (ch == ‘A‘ && a + t <= 60000)
29                     a += t;
30                 else if (ch == ‘B‘ && b + t <= 60000)
31                     b += t;
32                 else if (ch == ‘C‘ && c + t <= 60000)
33                     b += t;
34                 else
35                     flag = false;
36 
37             }
38             if (a + b + c <= 100000 && a <= 60000 && b <= 60000 && c <= 60000 && flag)
39                 money[len++] = a + b + c;//如果满足以上三个条件
40         }
41         for (int i = 0; i < len; i++)//01背包核心代码
42         {
43             for (int j = sum; j >= money[i]; j--)
44                 if (dp[j] < dp[j - money[i]] + money[i])
45                     dp[j] = dp[j - money[i]] + money[i];
46         }
47         printf("%.2f\n", (dp[sum]/100.0));
48     }
49     return 0;
50 }