2019牛客暑期多校训练营（第一场）A题【单调栈】（补题）

链接：https://ac.nowcoder.com/acm/contest/881/A
来源：牛客网

题目描述 

Two arrays u and v each with m distinct elements are called equivalent if and only if RMQ(u,l,r)=RMQ(v,l,r)RMQ(u,l,r)=RMQ(v,l,r) for all 1≤l≤r≤m1≤l≤r≤m
where RMQ(w,l,r)RMQ(w,l,r) denotes the index of the minimum element among wl,wl+1,…,wrwl,wl+1,…,wr.
Since the array contains distinct elements, the definition of minimum is unambiguous.

Bobo has two arrays a and b each with n distinct elements. Find the maximum number p≤np≤n where {a1,a2,…,ap}{a1,a2,…,ap} and {b1,b2,…,bp}{b1,b2,…,bp} are equivalent.

input

The input consists of several test cases and is terminated by end-of-file.

The first line of each test case contains an integer n.
The second line contains n integers ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an.The\; third\; line\; contains\; n\; integers\; b1,b2,\dots ,bnb1,b2,\dots ,bn.*\; 1\le n\le 1051\le n\le 105*\; 1\le ai,bi\le n1\le ai,bi\le n*\; \{a1,a2,\dots ,an\}\{a1,a2,\dots ,an\}\; are\; distinct.*\; \{b1,b2,\dots ,bn\}\{b1,b2,\dots ,bn\}\; are\; distinct.*\; The\; sum\; of\; n\; does\; not\; exceed\; 5\times 1055\times 105.OUTPUT}}^{}$

For each test case, print an integer which denotes the result.
输入

2
1 2
2 1
3
2 1 3
3 1 2
5
3 1 5 2 4
5 2 4 3 1

输出

1
3
4

题意：给出两个序列A和B，让你找出最大的一个p，使得在区间[1,p]内，任意的l，r∈[1,n],RMQ(l,r,A)要等于RMQ(l,r,B)。RMQ(l,r,A)返回的是[l,r]内最小值对应的下标。

思路：让${\mathrm{ai和bi同时进入单调队列，单调队列保存元素的值和下标，如果进队/出队的时候，ai所在的单调队列和bi所在的单调队列的下标值是一样的，那么就是OK的，直到找到状态不一样的位置，那么答案为它的前一个。渐进时间复杂度O(n)。}}^{}$AC代码：

#include<bits/stdc++.h>

using namespace std;
struct str{
    int val;    
        int pos;

};
int main(){
    int n;
    while(~scanf("%d",&n)){
        int a[n+10];
        int b[n+10];
        deque<str> q1,q2;
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)
            scanf("%d",&b[i]);
        q1.push_front((str){-10000000,0});
        q2.push_front((str){-10000000,0});
        int arr1[n+10];
        int arr2[n+10];
        for(int i=1;i<=n;i++){
            while(!q1.empty()&&q1.front().val>a[i])
                q1.pop_front();
            arr1[i]=q1.front().pos;
            q1.push_front((str){a[i],i});
            while(!q2.empty()&&q2.front().val>b[i])
                q2.pop_front();
            arr2[i]=q2.front().pos;
            q2.push_front((str){b[i],i});
        }
        int ans=1;
    
        for(int i=1;i<=n;i++){
            if(arr1[i]==arr2[i]){
                ans=i;
            }else{
                break;
            }
        }
        printf("%d
",ans);
    }
    
    return 0;
}