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题目链接：http://codeforces.com/problemset/problem/631/C
Report
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 manager i + 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 integers ti 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
3 1
1 2 3
2 2
2 1 3 4 2
1 2 4 3
2 3
1 2
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.
思路：首先，对于两个不同位置的操作，如果针对第i个操作，若有第j个操作(1<=j<i)且(xj<=xi)，则第j个操作是可以忽略不计的。这样，我们就只需要维护一个xi严格递增的单调栈，将所有有效操作筛选出来，然后在两个邻近的操作之间填数就可以了。详见代码。
附上AC代码：
#include <bits/stdc++.h>
#define pii pair<int, int>
#define f first
#define s second
using namespace std;
const int maxn = 200005;
int num[maxn], ans[maxn];
pii op[maxn];
int n, m;int main(){ios::sync_with_stdio(false);cin.tie(0);cin >> n >> m;for (int i=1; i<=n; ++i){cin >> num[i];ans[i] = num[i];}int top=0, x, y;for (int i=1; i<=m; ++i){cin >> x >> y;while (top && y>=op[top-1].s)--top;op[top].f=x, op[top].s=y;++top;}sort(num+1, num+op[0].s+1);int low=1, high=op[0].s;op[top].f = op[top].s = 0;for (int i=0; i<top; ++i)for (int j=op[i].s; j>op[i+1].s; --j){if (op[i].f == 1)ans[j] = num[high--];elseans[j] = num[low++];}for (int i=1; i<=n; ++i)cout << ans[i] << (i!=n ? ' ' : '\n');return 0;
}
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