CodeForces 689D Friends and Subsequences

Posted 2020-08-06 Fighting Heart

枚举，二分，$RMQ$。

对于一个序列来说，如果固定区间左端点，随着右端点的增大，最大值肯定是非递减的，最小值肯定是非递增的。

因此，根据这种单调性，我们可以枚举区间左端点$L$，二分找到第一个位置${{p_1}}$，使得$\mathop {\max }\limits_{i = L}^{{p_1}} {a_i} = \mathop {\min }\limits_{i = L}^{{p_1}} {b_i}$；再次二分找到最后一个位置${{p_2}}$，使得$\mathop {\max }\limits_{i = L}^{{p_2}} {a_i} = \mathop {\min }\limits_{i = L}^{{p_2}} {b_i}$。那么以$L$为左端点的区间，有${{p_2}}-{{p_1}}+1$个。查询区间最值的话可以倍增预处理一下。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-6;
void File()
{
    freopen("D:\\in.txt","r",stdin);
    freopen("D:\\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c = getchar();
    x = 0; while(!isdigit(c)) c = getchar();
    while(isdigit(c)) { x = x * 10 + c - ‘0‘; c = getchar(); }
}

const int maxn=200010;
int a[maxn],b[maxn],n;
int MAX[maxn][30],MIN[maxn][30];

void RMQ_init()
{
    for(int i=0;i<n;i++) MAX[i][0]=a[i],MIN[i][0]=b[i];
    for(int j=1;(1<<j)<=n;j++)
        for(int i=0;i+(1<<j)-1<n;i++)
            MAX[i][j]=max(MAX[i][j-1],MAX[i+(1<<(j-1))][j-1]),
            MIN[i][j]=min(MIN[i][j-1],MIN[i+(1<<(j-1))][j-1]);
}

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

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

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]);
    RMQ_init();

    LL ans=0;
    for(int i=0;i<n;i++)
    {
        if(b[i]<a[i]) continue;
        int p1=-1,p2=-1;
        int L=i,R=n-1;
        while(L<=R)
        {
            int mid=(L+R)/2;
            int mx=RMQ_MAX(i,mid),mn=RMQ_MIN(i,mid);
            if(mx>mn) R=mid-1;
            else if(mx==mn) R=mid-1,p1=mid;
            else L=mid+1;
        }

        L=i,R=n-1;
        while(L<=R)
        {
            int mid=(L+R)/2;
            int mx=RMQ_MAX(i,mid),mn=RMQ_MIN(i,mid);
            if(mx>mn) R=mid-1;
            else if(mx==mn) L=mid+1,p2=mid;
            else L=mid+1;
        }

        if(p1==-1) continue;
        ans=ans+(LL)(p2-p1+1);
    }
    printf("%lld\n",ans);

    return 0;
}