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

    D. Friends and Subsequences

    题目连接:

    http://www.codeforces.com/contest/689/problem/D

    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 is satisfied

    Sample Input

    6
    1 2 3 2 1 4
    6 7 1 2 3 2

    Sample Output

    2

    Hint

    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。

    代码

    #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[n]=2e9;
        b[n]=-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;
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/5651384.html
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