Codeforces Round #361 (Div. 2) D RMQ+二分

連結：戳這裡

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  is satisfied.

Examples
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
Note
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.

題意：

分別給出n個數ai和bi

在區間[l,r]中存在max(al~ar)==min(bl~br) 問這樣的區間有多少個

思路：

題目很經典，我們先用RMQ處理出區間[l,r]的最值，方便log去訪問

為什麼可以這樣，因為倍增越往後max肯定越大，同理min肯定越小，所以可以二分快速找

那麼我們固定區間[l,r]上的l，再去找符合要求的r

很容易發現滿足這個條件的r肯定也是一段距離

那麼什麼時候是滿足要求呢？

當列舉的區間為[i,r1] （i是定死的）內正好max(ai~ar1)==min(bi~br1) 表示這一段是滿足的 ->ans+1

那麼接下來從[r1~n]區間裡還能不能延長r1使得也滿足情況

那麼只需要找出一個r2使得這個區間的max(a)正好大於min(b)，這樣作為分界點去統計答案

程式碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
int n;
int a[200100],b[200100];
int rmq1[200100][20],rmq2[200100][20];
int query1(int l,int r){
    int m=log(r-l+1)/log(2);
    return max(rmq1[l][m],rmq1[r-(1<<m)+1][m]);
}
int query2(int l,int r){
    int m=log(r-l+1)/log(2);
    return min(rmq2[l][m],rmq2[r-(1<<m)+1][m]);
}
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    for(int i=1;i<=n;i++) scanf("%d",&b[i]);
    a[++n]=2e9;b[n]=-2e9;
    for(int i=1;i<=n;i++) rmq1[i][0]=a[i];
    for(int i=1;i<=n;i++) rmq2[i][0]=b[i];
    int m=log(n)/log(2);
    for(int i=1;i<=m;i++){
        for(int j=n;j>=1;j--){
            rmq1[j][i]=rmq1[j][i-1];
            if(j+(1<<(i-1))<=n)
                rmq1[j][i]=max(rmq1[j][i],rmq1[j+(1<<(i-1))][i-1]);
        }
    }
    for(int i=1;i<=m;i++){
        for(int j=n;j>=1;j--){
            rmq2[j][i]=rmq2[j][i-1];
            if(j+(1<<(i-1))<=n)
                rmq2[j][i]=min(rmq2[j][i],rmq2[j+(1<<(i-1))][i-1]);
        }
    }
    ll ans=0;
    for(int i=1;i<=n;i++){
        if(a[i]>b[i]) continue;
        int l=i,r=n,mid,L=-1;
        while(l<r){
            mid=(l+r)/2;
            if(query1(i,mid)>=query2(i,mid)){
                L=mid;
                r=mid;
            } else l=mid+1;
        }
        if(L==-1) continue;
        l=i;r=n;
        int R=i;
        while(l<r){
            mid=(l+r)/2;
            if(query1(i,mid)>query2(i,mid)){
                R=mid;
                r=mid;
            } else l=mid+1;
        }
        if(query1(i,l)>query2(i,l)) R=l;
        ans+=R-L;
    }
    printf("%I64d\n",ans);
    return 0;
}
/*
5
1 1 1 1 1
1 1 1 1 1
*/