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A.Equivalent Prefixes

The meaning of problems: Given two arrays containing n elements a, b, the array does not contain the same element, seeking the maximum p (p <= n), such that a and b are equivalent, an equivalent herein is defined as a [ l, r] and b [l, r] is equal to the minimum value of the subscript (1 <= l <= r <= p).

answer:

1, p ranges from 1 ~ n, from small to large to traverse, monotone stack get La [i], represents a ratio of empathy index a [i] smaller elements, Lb [i]. I-th element, if Lu [i]! = Lv [i], including a section comprising i described condition is not satisfied, then p can take the maximum i-1.

Monotonous stack

AC Code:

 1 #include<iostream>
 2 #include<sstream>
 3 #include<fstream>
 4 #include<algorithm>
 5 #include<string>
 6 #include<cstring>
 7 #include<iomanip>
 8 #include<cstdlib>
 9 #include<cctype>
10 #include<vector>
11 #include<cmath>
12 #include<ctime>
13 #include<stack>
14 #include<queue>
15 #include<map>
16 #include<set>
17 #define mem(a,b) memset(a,b,sizeof(a))
18 #define random(a,b) (rand()%(b-a+1)+a)
19 #define ull unsigned long long
20 #define e 2.71828182
21 #define Pi acos(-1.0)
22 using namespace std;
23 const int MAXN=1e5+5;
24 int n,a[MAXN],b[MAXN];
25 int La[MAXN],Lb[MAXN]; 
26 stack<int> sa,sb;
27 int read()
28 {
29     int s=1,x=0;
30     char ch=getchar();
31     while(!isdigit(ch)) {if(ch=='-') s=-1;ch=getchar();}
32     while(isdigit(ch)) {x=10*x+ch-'0';ch=getchar();}
33     return x*s;
34 }
35 void init()
36 {
37     while(!sa.empty()) sa.pop();
38     while(!sb.empty()) sb.pop();
39     for(int i=1;i<=n;++i)
40     {
41         while(!sa.empty()&&a[i]<a[sa.top()]) sa.pop();
42         if(sa.empty()) La[i]=0;
43         else La[i]=sa.top();
44         sa.push(i);
45     }
46     for(int i=1;i<=n;++i)
47     {
48         while(!sb.empty()&&b[i]<b[sb.top()]) sb.pop();
49         if(sb.empty()) Lb[i]=0;
50         else Lb[i]=sb.top();
51         sb.push(i);
52     }
53 }
54 int main()
55 {
56     ios::sync_with_stdio(false);
57     cin.tie(0);cout.tie(0);
58     while(~scanf("%d",&n))
59     {
60         getchar();
61         for(int i=1;i<=n;++i) a[i]=read();
62         for(int i=1;i<=n;++i) b[i]=read();
63         int ans=n;
64         init();
65         for(int i=2;i<=n;++i)
66         {
67             if(La[i]!=Lb[i])
68             {
69                 ans=i-1;
70                 break;
71             } 
72         }
73         cout<<ans<<endl;
74     }
75 }

View Code

2, Cartesian tree approach

 

 

Integration