UVa 12299 RMQ with Shifts（移位RMQ）

UVa 12299 - RMQ with Shifts（移位RMQ）

Time limit: 1.000 seconds

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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

 

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。

 

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。

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）。

 

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++

#include <cstdio>
#include <cstring>
#include <algorithm>
#define mx 100005
int tr[mx << 2], path[mx], iP, ts[mx];
int build(int L, int R, int now){
    if (L > R) return mx;
    if (L == R) scanf("%d", tr + (path[++iP] = now));
    else{
        int mid = L + R >> 1, cL, cR;
        cL = build(L, mid, now << 1); cR = build(mid + 1, R, now << 1 | 1);
        tr[now] = std::min(cL, cR);
    }
    return tr[now];
}
int query(int L, int R, int now, int edL, int edR){
    if (edL <= L && R <= edR) return tr[now];
    int mid = L + R >> 1, cL, cR;
    if (edR <= mid) return query(L, mid, now << 1, edL, edR);
    if (mid < edL) return query(mid + 1, R, now << 1 | 1, edL, edR);
    cL = query(L, mid, now << 1, edL, mid); cR = query(mid + 1, R, now << 1 | 1, mid + 1, edR);
    return std::min(cL, cR);
}
void updata(int now, int n){
    int tmp, cL, cR;
    while (now >>= 1){
        tmp = std::min(tr[now << 1], tr[now << 1 | 1]);
        if (tmp == tr[now]) return;
        tr[now] = tmp;
    }
}
int main(){
    memset(tr, 127, sizeof(tr));
    int n, q, i, j, tmp;
    char op[20];
    scanf("%d%d", &n, &q);
    build(1, n, 1); getchar();
    while (q--){
        fread(op, sizeof(char), 6, stdin);
        if (*op == 'q'){
            for (i = 0; i < 2; ++i) scanf("%d", ts + i), getchar();
            printf("%d
", query(1, n, 1, ts[0], ts[1]));
        }
        else{
            for (j = 0; *op != ')'; ++j) scanf("%d", ts + j), *op = getchar();
            for (tmp = tr[path[ts[i = 0]]]; i < j - 1; ++i){
                tr[path[ts[i]]] = tr[path[ts[i + 1]]];
                updata(path[ts[i]], n);
            }
            tr[path[ts[i]]] = tmp;
            updata(path[ts[i]], n);
        }
        getchar();
    }
    return 0;
}