# The so-called daily #4-the abolition of the Emperor Chan practices position thief mengde dedication Knife

Source: Internet
Author: User

Div.2

Codeforces 459A Pashmak and garden

There is a square on the two-dimensional plane, and its edge is parallel to the axis. Now give two points and ask if there is a square vertex that is these two points. Outputs the coordinates of the other two points, if any.

It would be nice to have a discussion on the side-length relationship.

`1#include <stdio.h>2#include <algorithm>3 4 intx[4],y[4];5 6 intMain () {7      for(inti =0; I <2; ++i) {8scanf"%d%d",&x[i],&y[i]);9     }Ten     intDX = x[0]-x[1]; One     intDY = y[0]-y[1]; A     if(dx!=0&& dy!=0&& std::abs (dx)! =std::abs (DY)) { -Puts"-1"); -         return 0; the     } -     if(dx==0) { -x[2] = x[0] +dy; -y[2] = y[0]; +x[3] = x[1] +dy; -y[3] = y[1]; +}Else if(dy==0) { Ax[2] = x[0]; aty[2] = y[0] +DX; -x[3] = x[1]; -y[3] = y[1] +DX; -}Else { -x[2] = x[0]; -y[2] = y[1]; inx[3] = x[1]; -y[3] = y[0]; to     } +printf"%d %d%d%d\n", x[2],y[2],x[3],y[3]); -     return 0; the}`
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Div.1

Codeforces 566D Restructuring Company

There is a row of n (<= 2e5) points that perform 5e5 operations. There are three kinds of operation: (1,x,y) connect x point with Y Point, (2,x,y) connect (x,x+1,x+2...y-1,y), (3,x,y) ask whether X point and Y point are connected.

It looks very, very well, and it's got wood.

Requires two and check sets, 1 maintenance connectivity, and 1 references to daily #1div1, and maintains only the right connecting point for each point under Operation 2.

Finally someone in the game time over the Div.1 of the problem, moved AH have wood.

`1#include <bits/stdc++.h>2 3 Const intN =200000+5;4 intn,m;5 intdsu[2][n];6 7 intFindint*DSU,intx) {8     returnDSU[X] = = x? X:DSU[X] =find (Dsu,dsu[x]);9 }Ten  One intMain () { Ascanf"%d%d",&n,&m); -      for(inti =0; I <= N; ++i) { -dsu[0][i] = dsu[1][i] =i; the     } -      while(m--) { -         intop,a,b; -scanf"%d%d%d", &op,&a,&b); A--; B--; +         if(OP = =1) { -             intx = Find (dsu[0],a); +             inty = Find (dsu[0],b); Adsu[0][X] =y; at}Else if(OP = =2) { -              for(inti = Find (dsu[1],A); I < b; dsu[1][i] = i +1, I = Find (dsu[1],i)) { -                 intx = Find (dsu[0],i); -                 inty = Find (dsu[0],i +1); -dsu[0][X] =y; -             } in}Else { -Puts (Find (dsu[0],a) = = = Find (dsu[0],B)?"YES":"NO"); to         } +     } -}`
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The so-called daily #4-the abolition of the Emperor Chan practices position thief mengde dedication Knife

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