P1375 nested rectangle and p1375 nested rectangle

P1375 nested rectangle and p1375 nested rectangle
QuestionProblem nested rectangle Time Limit: 1000 ms Memory Limit: 131072KBDescriptionDescript.

There are n rectangles. Each rectangle can be described by a and B to indicate length and width. Rectangle X (a, B) can be nested in Rectangle Y (c, d) When and only when a <c, B <d or B <c, a <d (equivalent to rotating x 90 degrees ). For example, () can be nested in (), but cannot be nested in. Your task is to select as many rectangles as possible in a row, so that, except the last one, each rectangle can be nested in the next rectangle.

InputInput 1st rows n (n <= 2000)
Two numbers a and B in each row from 2nd to n + 1 indicate the length and width of the rectangle. OutputOutput is a single number. The maximum number of rectangles that meet the condition is ExampleSample input data

31 56 23 4

Output Data

2

RemarksHint smartoj does not have an evaluation machine... I don't know, right ,,

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 using namespace std; 6 const int MAXN=2001; 7 void read(int & n) 8 { 9     char c='+';int x=0;bool flag=0;10     while(c<'0'||c>'9')11     {c=getchar();if(c=='-')flag=1;}12     while(c>='0'&&c<='9')13     {x=x*10+(c-48);c=getchar();}14     flag==1?n=-x:n=x;15 }16 int map[MAXN][MAXN];17 struct node18 {19     int hang;20     int lie;21     int id;22 }a[MAXN*4];23 int ans=0;24 int n;25 int dis[MAXN];26 int M_s(int p)27 {28     ans=max(ans,dis[p]);29     if(dis[p])30     return dis[p];31     for(int i=1;i<=n;i++)32     {33         if(map[p][i])34         return dis[p]=max(dis[p],M_s(i)+1);35     }36 }37 int main()38 {39     read(n);40     for(int i=1;i<=n;i++)41     {42         int x,y;43         read(x);read(y);44         a[i].hang=x;a[i].lie=y;a[i].id=i;45     }46     for(int i=1;i<=n;i++)47         for(int j=1;j<=n;j++)48             if(i!=j)49                 if((a[i].hang<a[j].hang&&a[i].lie<a[j].lie)||(a[i].lie<a[j].hang&&a[i].hang<a[j].lie))50                     map[a[i].id][a[j].id]=1;51     52     M_s(1);53     int out=1;54     for(int i=1;i<=n;i++)55     {56         out=max(out,dis[i]+1);57     }58     printf("%d",out);59     return 0;60 }