hiho 1068 重新整理的 Sparse-Table(RMQ)模板

http://hihocoder.com/problemset/problem/1067
代码:
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <cmath>
 6 #include <algorithm>
 7 #include <string>
 8 #include <queue>
 9 #include <stack>
10 #include <vector>
11 #include <map>
12 #include <set>
13 #include <functional>
14 #include <cctype>
15 #include <time.h>
16 
17 using namespace std;
18 
19 struct Sparse_Table {
20     static const int MAXN = (int)1e6+55;
21     static const int MAXM = 21;
22 
23     int ST[MAXM][MAXN];
24     int bit[MAXN];
25     int n;
26 
27     void init() { //初始化bit数组，放在程序开始的时候，只需做一次
28         bit[0] = bit[1] = bit[1] = 0;
29         for (int i = 1; (i<<1)<MAXN; i++)
30             bit[i<<1] = bit[i<<1|1] = bit[i]+1;
31     }
32 
33     void initTable(int a[], int n) {
34         this->n = n;
35         for (int i = 0; i < n; i++)
36             ST[0][i] = a[i];
37         int hight = bit[n];
38         for (int i = 0; i < hight; i++)
39             for (int j = 0; j+(1<<i) < n; j++)
40                     ST[i+1][j] = min(ST[i][j], ST[i][j+(1<<i)]);
41     }
42 
43     inline int Query(int l, int r) {
44         if (l==r) return ST[0][l];
45         int h = bit[r-l+1];
46         return min(ST[h][l], ST[h][r-(1<<h)+1]);
47     }
48 
49     void output() {
50         int h = bit[n];
51         for (int i = 0; i <= h; i++) {
52             for (int j = 0; j < n; j++)
53                 printf("%d ", ST[i][j]);
54             puts("");
55         }
56     }
57 };
58 
59 const int INF = 1<<30;
60 const int MAXN = 1e6+55;
61 
62 Sparse_Table solver;
63 int a[MAXN];
64 int n, Q;
65 
66 int main() {
67     #ifdef Phantom01
68         freopen("1068.txt", "r", stdin);
69     #endif //Phantom01
70 
71     solver.init();
72 
73     while (scanf("%d", &n)!=EOF) {
74         for (int i = 0; i < n; i++) scanf("%d", &a[i]);
75 
76         solver.initTable(a, n);
77         scanf("%d", &Q);
78         for (int i = 0; i < Q; i++) {
79             int l, r;
80             scanf("%d%d", &l, &r);
81             printf("%d
", solver.Query(l-1, r-1));
82         }
83     }
84 
85     return 0;
86 }
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