poj1787Charlie's Change(多重背包+记录路径+好题)

 1 #include <iostream>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <cstdio>
 5 
 6 using namespace std;
 7 const int MAX = 10000 + 10;
 8 int dp[MAX],num[MAX][5],c[5],p;
 9 int w[5]={1,5,10,25};
10 void ZeroOnePage(int cost, int mount, int kind) //01背包传递参数有这一种硬币的数量和种类
11 {
12     for(int i = p; i >= cost; i--)
13     {
14         if(dp[i - cost] > -1 && dp[i] < dp[i - cost] + mount) //求枚数最多的情况，01的数量需要参数传递，完全背包就是1
15         {
16             dp[i] = dp[i - cost] + mount;
17             for(int j = 0; j < 4; j++)
18                 num[i][j] = num[i - cost][j];  //更改这一状态下每一种硬币的数量
19             num[i][kind] += mount;    
20         }
21     }
22 }
23 void CompletePage(int cost, int kind)
24 {
25     for(int i = cost; i <= p; i++)
26     {
27         if(dp[i - cost] > -1 && dp[i] < dp[i - cost] + 1)
28         {
29             dp[i] = dp[i - cost] + 1;
30             for(int j = 0; j < 4; j++)
31                 num[i][j] = num[i - cost][j];
32             num[i][kind] += 1;
33         }
34     }
35 }
36 void MultiplePage(int cost,int mount,int kind)
37 {
38     if(cost * mount >= p)
39     {
40         CompletePage(cost, kind);
41         return;
42     }
43     int k = 1;
44     while(k < mount)
45     {
46         ZeroOnePage(k * cost, k, kind);
47         mount = mount - k;
48         k <<= 1;
49     }
50     if(mount > 0)
51         ZeroOnePage(mount * cost, mount, kind);
52     return ;
53 }
54 int main()
55 {
56     while(scanf("%d", &p) != EOF)
57     {
58         int sum = p;
59         for(int i = 0; i < 4; i++)
60         {
61             scanf("%d", &c[i]);
62             sum += c[i];
63         }
64         if(sum == 0)
65             break;
66         memset(num,0,sizeof(num));
67         memset(dp,-1,sizeof(dp));
68         dp[0] = 0;
69         for(int i = 0; i < 4; i++)
70             if(c[i])
71             MultiplePage(w[i],c[i],i);
72         if(dp[p] > 0)
73         {
74             printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n",num[p][0],num[p][1],num[p][2],num[p][3]);
75         }
76         else
77         {
78             printf("Charlie cannot buy coffee.\n");
79         }
80     }
81     return 0;
82 }