HDU-2844 Coins(多重背包)

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <algorithm>
 5 
 6 using namespace std;
 7 const int max_size = 1000000 + 10;
 8 const int MAX = 105;
 9 int dp[max_size];
10 bool vis[max_size];
11 
12 
13 
14 int main()
15 {
16     //1.将问题的模型抽象出来
17     int cnt, vol;
18     int val[MAX];
19     int num[MAX];
20     while(scanf("%d %d", &cnt, &vol) != EOF)
21     {
22         memset(dp, 0, sizeof(dp));
23         memset(vis, false, sizeof(vis));
24         if(cnt == 0 && vol == 0)
25             break;
26         for(int i = 0; i < cnt; i++)
27             scanf("%d", val+i);
28         for(int i = 0; i < cnt; i++)
29             scanf("%d", num+i);
30 
31         for(int i = 0; i < cnt; i++)
32         {
33             if(val[i] * num[i] >= vol)
34             {
35                 //CompletePack(val[i], val[i]); //那么多的价值，那么多的占用？
36 
37                 for(int j = val[i]; j <= vol; j++)      ///多重背包这里错了两次了，要注意，昨天找了一晚上
38                 {
39                     dp[j] = max(dp[j], dp[j - val[i]] + val[i]);
40                     vis[dp[j]] = true;
41                 }
42                 continue;
43             }
44             int k = 1;
45             while(k < num[i])
46             {
47                 //ZeroOnePack(k*val[i], k*val[i]);
48                 for(int j = vol; j - k * val[i] >= 0; j--)
49                 {
50                     dp[j] = max(dp[j], dp[j-k*val[i]] + k*val[i]);
51                     vis[dp[j]] = true;
52                 }
53                 num[i] -= k;
54                 k *= 2;
55             }
56             //ZeroOnePack(num[i]*val[i], num[i]*val[i]);
57             for(int j = vol; j - num[i]*val[i] >= 0; j--)
58             {
59                 dp[j] = max(dp[j], dp[j - num[i]*val[i]] + num[i]*val[i]);
60                 vis[dp[j]] = true;
61             }
62         }
63 
64         int ans = 0;
65         for(int i = 1; i <= vol; i++)
66         {
67             ans += (vis[i] == true) ? 1 : 0;
68         }
69         printf("%d\n", ans);
70     }
71     return 0;
72 }