Tree-like array-the solving of artificial lake

Title: Artificial Lake

Describe:

"Title description"

There is a lake, surrounded by cities, and each city is connected to only two cities adjacent to him. Suppose there are N cities, numbered 1-n, highways are bidirectional, highways are sometimes good, sometimes bad, and now ask you whether two cities can reach each other.

"Input Format"

The first row is two numbers, one 2<=n<=100000 and 1<=m<=100000, each representing the number of cities and the number of queries; the next M-line, three-digit f,a,b per line. F=0, if the road between a A and a is good before, now it becomes bad, if it is bad, now it becomes good. When f=1, ask if the two cities of a, B can reach each other.

"Output Format"

For each f=1 query, the output "YES" can be reached, otherwise the output "NO".

"Sample Input"5 Ten

1 2 5

0 4 5

1 4 5

0 2 3

1 3 4

1 1 3

0 1 2

0 2 3

1 2 4

1 2 5

"Sample Output"

YES

YES

YES

NO

YES

NO

Prompted

30% 2<=n,m<=100

50% 2<=n,m<=10000

100% 2<=n,m<=100000

At first glance, it is easy to think of a search can be used, but because M is too large, plus the range of n is not small, the complexity of the maximum $o (NM) $, so 30 points abandoned therapy ... However, a tree-like array can be a good solution to this problem. Open the array to 2*n, subscript 1. 2*n, the X-cell represents the path state from the X-to (x+1) mod 2*n, which is indicated by 1, and 0 means no. Open an array to store the sum of the roads in some intervals, then make a judgment on each change, and then change the values in the tree array two times. In the case of a query operation, it is calculated two times:

1. Seek the total state of the path between the x~y and then judge (note that if it is ' no ', do not jump out)

2. Seek the total state of the path between the y~x+n and then judge it.

The implementation of the concrete is not difficult, the middle of the addition of some small tricks. But...... F=0, the size of x, Y is not necessarily small to large ... I was all over WA once ... Change it again before AC ... In a word, I finally see a tree-like array of questions that are not so fruit ...

AC Code:

{

Program zht;vari,j,k,m,n,y,x,f,t,z1,z2:longint;a:array[0..200000] of longint;c:array[0..200000] of Longint;   Procedure Change;vark:longint;begink:=0;if a[x]=1 then BEGIN t:=-1;   a[x]:=0;   a[x+n]:=0;    end ELSE begin t:=1;     A[x]:=1;     A[x+n]:=1;                    End Change the state of k:=x; While K<=2*n does begin c[k]:=c[k]+t; K:=k+ (k and (k)); End;k:=x+n;while k<=2*n do begin c[k]:=c[k]+t; K:=k+ (k and (k));                                  End       Tree-like array two-time change end; Gai Bian Zhuang taiprocedure chazhao;vark:longint;begink:=0;k:=x-1; While k>0 do begin z1:=z1+c[k]; K:=k-(k and (k)); End;k:=0;k:=y-1; While k>0 do begin z2:=z2+c[k]; K:=k-(k and (k)); End;z1:=z2-z1;if Z1=y-x THEN BEGIN Writeln (' YES '); Exit        End Directly find z1:=0;z2:=0;k:=y-1;  While k>0 do begin z1:=z1+c[k];  K:=k-(k and (k)); end;k:=x+n-1;  While k>0 do begin z2:=z2+c[k];  K:=k-(k and (k)); End;if z2-z1=n-(y-x) then Writeln (' YES ') Else writeln (' NO '); Reverse two lookup end;beginassign (input, ' lakee.in '); Assign (output, ' lakee.out '); reset (input); rewrite (output); Readln (n,m); For I:=1 to 2*n-1 doa[i]:=1;for i:=1 to 2*n-1 dobegint:=i and (-i); For j:=i-t+1 to I does c[i]:=c[i]+1;end;for i:=1 to M dobeginreadln (f,x,y), t:=0;z1:=0;z2:=0;if x>y then begin t:=x; X:=y; y:=t;        End                        Pit Point ~~~t:=0;if f=0 then change else chazhao;end;                   The classification discusses close (input), close (output), and end. A short main program
}

<marvolo Original, non-reprint >

Tree-like array-the solving of artificial lake