A Equivalent Prefixes

原创于 2020-07-11 17:40:50 发布 · 254 阅读

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该博客讨论了在两个数组中，当它们具有相同的最小区间查询（RMQ）时，如何找到最大长度的等价前缀。问题描述了一个包含n个不同元素的数组a和b，目标是找到最长的子数组p，使得a的前p个元素和b的前p个元素等价。博客提供了一个示例输入和输出，并暗示可能包含解题代码。

链接：https://ac.nowcoder.com/acm/contest/6362/A
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)\mathrm{RMQ}(u, l, r) = \mathrm{RMQ}(v, l, r)RMQ(u,l,r)=RMQ(v,l,r) for all 1≤l≤r≤m1 \leq l \leq r \leq m1≤l≤r≤m
where RMQ(w,l,r)\mathrm{RMQ}(w, l, r)RMQ(w,l,r) denotes the index of the minimum element among wl,wl+1,…,wrw_l, w_{l + 1}, \dots, w_{r}wl,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 \leq np≤n where {a1,a2,…,ap}\{a_1, a_2, \dots, a_p\}{a1,a2,…,ap} and {b1,b2,…,bp}\{b_1, b_2, \dots, b_p\}{b1,b2,…,bp} are equivalent.

输入描述:

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 a1,a2,…,ana_1, a_2, \dots, a_na1,a2,…,an.
The third line contains n integers b1,b2,…,bnb_1, b_2, \dots, b_nb1,b2,…,bn.

* 1≤n≤1051 \leq n \leq 10^51≤n≤105
* 1≤ai,bi≤n1 \leq a_i, b_i \leq n1≤ai,bi≤n
* {a1,a2,…,an}\{a_1, a_2, \dots, a_n\}{a1,a2,…,an} are distinct.
* {b1,b2,…,bn}\{b_1, b_2, \dots, b_n\}{b1,b2,…,bn} are distinct.
* The sum of n does not exceed 5×1055 \times 10^55×105.

输出描述:

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

示例1

输入

复制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

复制1 3 4

1
3
4

代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int mod=1e9+7;
const int N=1e5+10;
ll c[N],n;
struct node{
    int id;
    int x;
    int l;
    int r;
}b[N],a[N];
bool cmp(node p,node q)
{
    return p.x<q.x;
}
bool bmp(node p,node q)
{
    return p.id<q.id;
}
int main()
{
    while(~scanf("%lld",&n))
    {
        for(int i=1;i<=n;i++)
        {
            cin>>a[i].x;
            a[i].id=i;
        }
        for(int i=1;i<=n;i++)
        {
            cin>>b[i].x;
            b[i].id=i;
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n+1;j++)
            {
                if(a[i].x>a[j].x||j==n+1)
                {
                    a[i].r=j-1;
                    break;
                }
            }
            for(int j=i-1;j>=0;j--)
            {
                if(a[i].x>a[j].x||j==0)
                {
                    a[i].l=j+1;
                    break;
                }
            }
            for(int j=i+1;j<=n+1;j++)
            {
                if(b[i].x>b[j].x||j==n+1)
                {
                    b[i].r=j-1;
                    break;
                }
            }
            for(int j=i-1;j>=0;j--)
            {
                if(b[i].x>b[j].x||j==0)
                {
                    b[i].l=j+1;
                    break;
                }
            }
        }
        int ans=min(a[1].r,b[1].r);
        for(int i=2;i<=n;i++)
        {
            if(i<=ans)
            {
                ans=min(ans,min(a[i].r,b[i].r));
            }
            else
            {
                if(a[i].l==b[i].l)
                ans=max(ans,min(a[i].r,b[i].r));
                else
                    break;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}