cf344C. Report

C. Report
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Each month Blake gets the report containing main economic indicators of the company “Blake Technologies”. There are n commodities produced by the company. For each of them there is exactly one integer in the final report, that denotes corresponding revenue. Before the report gets to Blake, it passes through the hands of m managers. Each of them may reorder the elements in some order. Namely, the i-th manager either sorts first ri numbers in non-descending or non-ascending order and then passes the report to the manageri + 1, or directly to Blake (if this manager has number i = m).
Employees of the “Blake Technologies” are preparing the report right now. You know the initial sequence ai of length n and the description of each manager, that is value ri and his favourite order. You are asked to speed up the process and determine how the final report will look like.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 200 000) — the number of commodities in the report and the number of managers, respectively.
The second line contains n integers ai (|ai| ≤ 109) — the initial report before it gets to the first manager.
Then follow m lines with the descriptions of the operations managers are going to perform. The i-th of these lines contains two integersti and ri ( , 1 ≤ ri ≤ n), meaning that the i-th manager sorts the first ri numbers either in the non-descending (if ti = 1) or non-ascending (if ti = 2) order.
Output
Print n integers — the final report, which will be passed to Blake by manager number m.
Examples
input
3 1
1 2 3
2 2
output
2 1 3
input
4 2
1 2 4 3
2 3
1 2
output
2 4 1 3
Note
In the first sample, the initial report looked like: 1 2 3. After the first manager the first two numbers were transposed: 2 1 3. The report got to Blake in this form.
In the second sample the original report was like this: 1 2 4 3. After the first manager the report changed to: 4 2 1 3. After the second manager the report changed to: 2 4 1 3. This report was handed over to Blake.

题意：给一个n大小数组，m次操作，每次操作对前r个数进行升序或降序操作。打印最后数组。
题解：首先跟操作的输入顺序有关。对于两个操作区间 ri ≥ rj, i > j,可以忽略rj 。这样就得到一个关于r的递减数组r[N]。

但是对每个r[i]都要排序一次吗？果断超时！首先对a[ 0…r[0] ]升序排序。然后确定a[r[i-1]…r[i] ]的值（从a[ 0…r[0] ]中选取合适的值）（因为a[r[i-1]…r[i] ]的值确定之后，后面对区间的排序都不会再考虑a[r[i-1]…r[i] ]的值），再确定a[r[i-2]…r[i-1] ]的值（从a[ 0…r[0] ]中选取合适的值）（因为a[r[i-2]…r[i-1] ]的值确定之后，后面对区间的排序都不会再考虑a[r[i-2]…r[i-1] ]的值）……这样直到a[ 0…r[0] ]的值都选完，只需一次排序。最后还要对a[0…r[n] ]进行排序。

#include <cstdio>#include <algorithm>const int N = (int)2e5 + 1;int n, m， a[N], ti, ri， t[N], r[N], s，b[N], bl, br;int main() {    scanf("%d%d", &n, &m), s = 0;    for (int i = 0; i < n; ++i) scanf("%d", &a[i]);    for (int i = 0; i < m; ++i) {        scanf("%d%d", &ti, &ri);        while (s > 0 && ri >= r[s - 1]) --s;        t[s] = ti, r[s] = ri, ++s;//s是ri的大小顺序     }    bl = 0, br = r[0], r[s] = 0, ++s;//     for (int i = 0; i < br; ++i)b[i] = a[i];    std::sort(b, b + br);    for (int i = 1; i < s; ++i)        for (int j = r[i - 1]; j > r[i]; --j)//确定a[r[i-1...r[i] ]的值             a[j - 1] = (t[i - 1] == 1) ? b[--br] : b[bl++];    for (int i = 0; i < n; ++i) printf("%d ", a[i]);    return 0;}