ST表与二分 （CodeForces 689D-Friends and Subsequences）

Friends and Subsequences

题目链接：

http://acm.hust.edu.cn/vjudge/contest/121333#problem/H

Description

Mike and !Mike are old childhood rivals, they are opposite in everything they do, except programming. Today they have a problem they cannot solve on their own, but together (with you) — who knows?

Every one of them has an integer sequences a and b of length n. Being given a query of the form of pair of integers (l, r), Mike can instantly tell the value of while !Mike can instantly tell the value of .

Now suppose a robot (you!) asks them all possible different queries of pairs of integers (l, r)(1 ≤ l ≤ r ≤ n) (so he will make exactly n(n + 1) / 2 queries) and counts how many times their answers coincide, thus for how many pairs is satisfied.

How many occasions will the robot count?

Input

The first line contains only integer n (1 ≤ n ≤ 200 000).

The second line contains n integer numbers a1, a2, ..., an ( - 109 ≤ ai ≤ 109) — the sequence a.

The third line contains n integer numbers b1, b2, ..., bn ( - 109 ≤ bi ≤ 109) — the sequence b.

Output

Print the only integer number — the number of occasions the robot will count, thus for how many pairs max(al~ar)==min(bl~br) is satisfied.

Sample Input

Input
6
1 2 3 2 1 4
6 7 1 2 3 2
Output
2
Input
3
3 3 3
1 1 1
Output
0

Hint

The occasions in the first sample case are:

1.l = 4,r = 4 since max{2} = min{2}.

2.l = 4,r = 5 since max{2, 1} = min{2, 3}.

There are no occasions in the second sample case since Mike will answer 3 to any query pair, but !Mike will always answer 1.

题意：

分别已知a b数组；
求有多少区间[l,r]满足max(al~ar)==min(bl~br);

分析：

首先建立ST表，便于计算任意区间的最大值和最小值。要想确定满足条件的区间个数，区间组合共（n-1）n/2 种，计算所有区间是显然超时的。

将区间的左端点l先固定住，右端点用二分求出满足条件的最小值first和最大值last，则对应于左端点为l的满足条件的区间个数为(last-first+1)，遍历左端点 l

//RMQ与二分#include <iostream>#include <cmath>using namespace std;int dmax[200005][20];int dmin[200005][20];int a[200005];int b[200005];void rmq_st(int n){    int i,j;    for(i=0;i<n;i++)        dmax[i][0]=a[i],dmin[i][0]=b[i];    int m=(int)(log(1.0*n)/log(2.0));    for(j=1;j<=m;j++)    {        for(i=0;i+(1<<j)-1<n;i++)        {            dmax[i][j]=max(dmax[i][j-1],dmax[i+(1<<(j-1))][j-1]);            dmin[i][j]=min(dmin[i][j-1],dmin[i+(1<<(j-1))][j-1]);        }    }}int rmq_findmax(int l,int r){  int k=(int)(log(1.0*(r-l+1))/log(2.0));    return max(dmax[l][k],dmax[r-(1<<k)+1][k]);}int rmq_findmin(int l,int r){    int k=(int)(log(1.0*(r-l+1))/log(2.0));    return min(dmin[l][k],dmin[r-(1<<k)+1][k]);}int main(){      int n,i,j;      cin>>n;      for(i=0;i<n;i++)          cin>>a[i];      for(i=0;i<n;i++)          cin>>b[i];      rmq_st(n);      long long ans=0;      for(i=0;i<n;i++)      {           int l=i,r=n-1,mid,flag=0,x,y,first,last;           //二分求右端点的最大值          while(l<=r)           {               mid=(l+r)/2;               x=rmq_findmax(i,mid);y=rmq_findmin(i,mid);               if(x==y)               {                   flag=1;                   last=mid;                        }               if(x>y)                //注意这是if而不是else if，目的是发现x==y时仍然向后查找，直到找到满足条件的最大值                    r=mid-1;               else l=mid+1;           }           //最终二分得到的结果的位置在r处          if(flag)           {               l=i;               r=n-1;               while(l<=r)                     //二分求右端点的最小值               {                   mid=(l+r)/2;                   x=rmq_findmax(i,mid);y=rmq_findmin(i,mid);                   if(x<y)                     l=mid+1;                   else r=mid-1,first=mid;          //当x==y时，仍然向前查找，直到找到满足条件的最小值               }               //最终二分得到正确结果的位置是在l 处               ans+=(last-first+1);           }      }      cout<<ans<<endl;    return 0;}