hduacm 3183 rmq

http://acm.hdu.edu.cn/showproblem.php?pid=3183

问题等价与取N-M个数，每次取的时候保证后面能取的个数足够，并且取的数最小查询最小用rmq

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<cstring>
 5 
 6 using namespace std;
 7 
 8 
 9 const int maxn = 10000 + 5;
10 
11 int N,M,d[maxn][20];
12 char s[maxn];
13 
14 
15 struct RMQ{
16     void init()
17     {
18         N = strlen(s);
19         for (int i = 0;i<N;i++)
20         d[i][0] = s[i];
21         for (int k = 1;(1<<k)<=N;k++)
22         {
23             for (int i = 0;i<N;i++)
24             d[i][k] = min(d[i][k-1],d[i+(1<<(k-1))][k-1]);
25 
26         }
27     }
28     int query(int L,int R)
29     {
30         int k = 0;
31         while (1<<(k+1)<=R-L+1) k++;
32         return min(d[L][k],d[R-(1<<k)+1][k]);
33     }
34 };
35 
36 RMQ rmq;
37 
38 int main()
39 {
40     while (~scanf("%s%d",s,&M))
41     {
42         rmq.init();
43         M = N - M;
44         int mv = 0;
45         bool flag = true;
46         for (int i = M;i;i--)
47         {
48             int c = rmq.query(mv,N-i);
49             while (mv<N&&s[mv]!=c) mv++;
50             mv++;
51             if (c=='0'&&flag)
52             continue;
53             flag = false;
54             printf("%c",c);
55 
56         }
57         if (flag)
58         printf("0");
59         printf("
");
60     }
61     return 0;
62 }

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