C. Report
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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 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
inputCopy
3 1
1 2 3
2 2
outputCopy
2 1 3
inputCopy
4 2
1 2 4 3
2 3
1 2
outputCopy
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：
给你一个长度为n的数组，和m个操作，一个操作有两种类型，1类型是将1到r的进行不下降排序，2类型是将1到r进行不上升排序。
思路：
如果直接按照题目模拟时间复杂度是 O(mn log(n) ) 显然会TLE的，那么我们考虑优化。
我们知道，如果一个操作是op1,r1，而后面有一个操作 op2，r2， 当r2 >= r1的时候，前一个操作显然是无意义的。那么我们可以用一个严格递减的单调栈来维护有意义的操作，然后把能改变到的数字放到一个multiset里，我们按照r从大到小处理，每一次只处理和上一个r的差值区间即可，通过multiset可以访问一段数字中最大值和最小值来分类处理不同的操作。
细节见代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int n,m;
int a[maxn];
stack<pii> b;
multiset<int> s;
std::vector<pii> v;
int main()
{
//freopen("D:\\code\\text\\input.txt","r",stdin);
//freopen("D:\\code\\text\\output.txt","w",stdout);
gbtb;
cin>>n>>m;
repd(i,1,n)
{
cin>>a[i];
}
int op,r;
int q=0;
repd(i,1,m)
{
cin>>op>>r;
while(!b.empty()&&b.top().se<=r)
{
b.pop();
}
b.push(mp(op,r));
q=max(q,r);
}
repd(i,1,q)
{
s.insert(a[i]);
}
v.push_back(mp(0,0));
while(!b.empty())
{
v.push_back(b.top());
b.pop();
}
int last=q;
for(int j=sz(v)-2;j>=0;--j)
{
op=v[j+1].fi;
for(int i=last;i>=v[j].se+1;i--)
{
if(op==1)
{
auto it=s.end();
it--;
a[i]=*it;
s.erase(it);
}else
{
a[i]=(*(s.begin()));
s.erase(s.begin());
}
}
last=v[j].se;
}
repd(i,1,n)
{
cout<<a[i]<<" ";
}
cout<<endl;
return 0;
}
inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
}
else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}
原文地址：https://www.cnblogs.com/qieqiemin/p/11336402.html