Codeforces Round #361 (Div. 2) D. Friends and Subsequences 二分

题意

给你一个a数组和一个b数组

问你有多少对(l,r)满足，a数组中max(L,R)恰好等于b数组中的min(L，R)

题解

暴力枚举L，然后二分相等的那个区间就好了。

因为max肯定是递增的，min是递减的

那个相等的区间可以二分出来。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5+7;
int n;
int a[maxn],b[maxn];
struct RMQ{
const static int RMQ_size = maxn;
int n;
int ArrayMax[RMQ_size][21];
int ArrayMin[RMQ_size][21];

void build_rmq(){
for(int j = 1 ; (1<<j) <= n ; ++ j)
for(int i = 0 ; i + (1<<j) - 1 < n ; ++ i){
ArrayMax[i][j]=max(ArrayMax[i][j-1],ArrayMax[i+(1<<(j-1))][j-1]);
ArrayMin[i][j]=min(ArrayMin[i][j-1],ArrayMin[i+(1<<(j-1))][j-1]);
}
}

int QueryMax(int L,int R){
int k = 0;
while( (1<<(k+1)) <= R-L+1) k ++ ;
return max(ArrayMax[L][k],ArrayMax[R-(1<<k)+1][k]);
}

int QueryMin(int L,int R){
int k = 0;
while( (1<<(k+1)) <= R-L+1) k ++ ;
return min(ArrayMin[L][k],ArrayMin[R-(1<<k)+1][k]);
}

void init(int * a,int sz){
n = sz ;
for(int i = 0 ; i < n ; ++ i) ArrayMax[i][0] = ArrayMin[i][0] = a[i];
build_rmq();
}

}s1,s2;

int main()
{
scanf("%d",&n);
for(int i=0;i<n;i++)scanf("%d",&a[i]);
for(int i=0;i<n;i++)scanf("%d",&b[i]);
a
=2e9;
b
=-2e9;
s1.init(a,n+1);
s2.init(b,n+1);
long long ans = 0;
for(int i=0;i<n;i++){
if(a[i]>b[i])continue;
int l=i,r=n,ansl=i;
while(l<=r){
int mid=(l+r)/2;
if(s1.QueryMax(i,mid)>=s2.QueryMin(i,mid))r=mid-1,ansl=mid;
else l=mid+1;
}
if(s1.QueryMax(i,ansl)>s2.QueryMin(i,ansl))continue;
l=i,r=n;
int ansr=i;
while(l<=r){
int mid=(l+r)/2;
if(s1.QueryMax(i,mid)>s2.QueryMin(i,mid))r=mid-1,ansr=mid;
else l=mid+1;
}
ans+=ansr-ansl;
}
cout<<ans<<endl;
}