Monotonic Queue: Temperature

The main idea: a country carried out a continuous n-day temperature measurement, measuring the existence of errors, the measurement result is the first day temperature within the [L_i,r_i] range.
The longest continuous paragraph satisfies the possibility that the temperature does not drop.
First row n
The following n lines, each line L_i,r_i
1<=n<=1000000
A row that represents the length of the segment
Sample Input
6
6 10
1 5
4 8
2 5
6 8
3 5
Sample Output
4

Knowledge points: recursive, monotone queue

1#include <bits/stdc++.h>2 using namespacestd;3queue<int>Q;4 intl[1000005],r[1000005];5 intq[1000005];6 intn,t;7 intMax,ans,head,tail;8 intMain () {9     //freopen ("temperature.in", "R", stdin);Ten     //freopen ("Temperature.out", "w", stdout); OneCin>>N; A      for(intI=1; i<=n;i++) -scanf"%d%d",&l[i],&r[i]); -     /* the monotonic Queue Analysis: - 1. What is the time of the break: when you want to join the first day, its r[i]<l[j] (L in the last few days), this time, - to update the answer first, and then re-count the day after the first day of my death - 2. Operating with a non-descending monotone queue, there is a key: Assume that the L value for a certain period of time is + 36 35 34 32 29 The following day L value is 45 This queue is updated to 55 53 48 45 ... Because the temperature - the number of days back and L value is high, if this day can be, then the first six days must be, so the queue element is a. + but this does not work, can not determine the number of days information, so let q[] save the day A L[q[head]] to call, update Q is to update the L array, anyway is at O (n) to sweep back from the past without affecting the answer -     */ -Head=1; tail=0; -      for(intI=1; i<=n;i++){ -          while(L[q[head]]>r[i]&&head<=tail)//Q is the monotone minus queue, if the current L[q[head]]>r[i] -head++//The day cannot be connected from q[head], so continue backwards. in                                          //find, until L[q[head]]<r[i], because it is a monotone queue, so -                                          //The next few days will be sure to          +     if(Head<=tail)//Update ans -Ans=max (ans,i-q[head]+1); the              *         intt=i; $      while(l[i]>l[q[tail]]&&head<=tail)Panax Notoginsengt=q[tail],tail--; -l[t]=L[i]; theq[++tail]=T; +     } Acout<<ans; the     return 0; + } -  

Monotonic Queue: Temperature