UVa 12299 - RMQ with Shifts（移位RMQ）
Time limit: 1.000 seconds
Description - 题目描述
In the traditional RMQ (Range Minimum Query) problem, we have a static array A. Then for each query (L,R) (L ≤ R), we report the minimum value among A[L], A[L + 1], ..., A[R]. Note that the indices start from 1, i.e. the left-most element is A[1]. In this problem, the array A is no longer static: we need to support another operation
在经典的RMQ（区间最值）问题中，给定数组A。对于每个询问(L,R) (L ≤ R)，输出A[L]， A[L + 1]， ...， A[R]中的最小值。注意下标从1开始，即最左边的元素为A[1]。在这个问题中，数组A会发生些许变化：定义如下操作
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shift(i1, i2, i3, ...,ik) (i1 < i2 < ... < ik, k > 1)
we do a left “circular shift” of A[i1], A[i2], ..., A[ik]. For example, if A={6, 2, 4, 8, 5, 1, 4}, then shift(2,4,5,7) yields {6, 8, 4, 5, 4, 1, 2}. After that,shift (1,2) yields 8, 6, 4, 5, 4, 1, 2.
我们对A[i1], A[i2], ..., A[ik]执行循环左移。例如，A={6, 2, 4, 8, 5, 1, 4}，则经过shift(2,4,5,7)后得到 {6, 8, 4, 5, 4, 1, 2}。再经过shift (1,2)得到8, 6, 4, 5, 4, 1, 2。
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Input - 输入
There will be only one test case, beginning with two integers n, q (1 ≤ n ≤ 100,000, 1 ≤ q ≤ 250,000), the number of integers in array A, and the number of operations. The next line contains n positive integers not greater than 100,000, the initial elements in array A. Each of the next q lines contains an operation. Each operation is formatted as a string having no more than 30 characters, with no space characters inside. All operations are guaranteed to be valid.
Warning: The dataset is large, better to use faster I/O methods.
只有一组测试用例，起始位置有两个整数n, q (1 ≤ n ≤ 100,000, 1 ≤ q ≤ 250,000)，分别表示整数数组A中的元素个数，询问的数量。下一行有n个不超过100,000的非负数，皆为数组A中的初始元素。随后q行每行包含一个操作。每个操作皆为一个不超过30个字符的字符串，且不含空格。全部操作均正确有效。 注意：数据量很大，最好使用更快的I/O函数。
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Output - 输出
For each query, print the minimum value (rather than index) in the requested range.
对于每个询问，输出待求范围的最小值（非下标）。
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Sample Input - 输入样例
7 5 6 2 4 8 5 1 4 query(3,7) shift(2,4,5,7) query(1,4) shift(1,2) query(2,2)
Sample Output - 输出样例
1 4 6
题解
一般的线段树。
虽然说数据量大，意思也就不能用cin吧，scanf还是可以A的。
一开始还以为需要用延迟更新，然后想了想30*25W的O(NlogN)还是应该可以的，能简则简。
代码中用的是自下往上的更新方式，目测比从上往下略快，可以跳过部分更新。
代码 C++
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #define mx 100005
5 int tr[mx << 2], path[mx], iP, ts[mx];
6 int build(int L, int R, int now){
7 if (L > R) return mx;
8 if (L == R) scanf("%d", tr + (path[++iP] = now));
9 else{
10 int mid = L + R >> 1, cL, cR;
11 cL = build(L, mid, now << 1); cR = build(mid + 1, R, now << 1 | 1);
12 tr[now] = std::min(cL, cR);
13 }
14 return tr[now];
15 }
16 int query(int L, int R, int now, int edL, int edR){
17 if (edL <= L && R <= edR) return tr[now];
18 int mid = L + R >> 1, cL, cR;
19 if (edR <= mid) return query(L, mid, now << 1, edL, edR);
20 if (mid < edL) return query(mid + 1, R, now << 1 | 1, edL, edR);
21 cL = query(L, mid, now << 1, edL, mid); cR = query(mid + 1, R, now << 1 | 1, mid + 1, edR);
22 return std::min(cL, cR);
23 }
24 void updata(int now, int n){
25 int tmp, cL, cR;
26 while (now >>= 1){
27 tmp = std::min(tr[now << 1], tr[now << 1 | 1]);
28 if (tmp == tr[now]) return;
29 tr[now] = tmp;
30 }
31 }
32 int main(){
33 memset(tr, 127, sizeof(tr));
34 int n, q, i, j, tmp;
35 char op[20];
36 scanf("%d%d", &n, &q);
37 build(1, n, 1); getchar();
38 while (q--){
39 fread(op, sizeof(char), 6, stdin);
40 if (*op == ‘q‘){
41 for (i = 0; i < 2; ++i) scanf("%d", ts + i), getchar();
42 printf("%d\n", query(1, n, 1, ts[0], ts[1]));
43 }
44 else{
45 for (j = 0; *op != ‘)‘; ++j) scanf("%d", ts + j), *op = getchar();
46 for (tmp = tr[path[ts[i = 0]]]; i < j - 1; ++i){
47 tr[path[ts[i]]] = tr[path[ts[i + 1]]];
48 updata(path[ts[i]], n);
49 }
50 tr[path[ts[i]]] = tmp;
51 updata(path[ts[i]], n);
52 }
53 getchar();
54 }
55 return 0;
56 }