Problem Link:
author: Tasnim Imran Sunny
tester: Roman Rubanenko
editorialist: Utkarsh Lath
Difficulty:
Medium-Hard
Pre-requisites:
ad-hoc
Problem:
The chef has written DFS traversal algorithm for R * C grid. He starts at (sr, sc) and terminates when he reaches (tc, tr). His algorithm looks for empty adjacent cells in the following order: right, down, left, up. You find need to find how many vertices he traverses before hitting (tc, tr).
Explanation:
In the most general case, the algorithm will proceed in one of the following ways depending on parity of (R-sr).
- For the case when parity of R-sr is even is easy to handle(refer to image on left), because the resulting pattern is very simple.
if sr == tr and sc >= tc
return tc-rc+1 // target to the immediate right of source
if sr < tr
if tc == C // target below source in last column
return (C-tc+1) + (tr-sr) // horizontal distance + vertical distance
return (C-sc+1) + (R-sr) + pattern1(C-1, R-tr, C-1-tc)
int inc = C - sc + 1 + C * (R-sr) // from DFS of part below source
if (R-cr) is even
if sr==tr // immediately to left of source
return inc + sc-tc
return inc + sc-1 + pattern1(C, sr-tr-1, tc-1) // above source
int pattern1(int cols, int top, int right) {
/ * solve for pattern
.
.
^-----<-------<-----
-->------>---------^
^-----<-------<-----
-->------>---------^
assuming you enter from bottom left, and move along direction suggested by arrows.
cols is number of columns
top is y-coordinate(0 based), right is x co-ordinate(0 based) * /
int res = cols * (top/2) * 2;
if top is odd
return res + (cols * 2) - right; // encountered when going from right to left
return res + right+1; // encountered when going from left to right
}
- In the other case(refer to image on right), the pattern is more interesting. It can end in one of the following ways.
1
2
3
4
//handling of cases when **sr == tr and sc >= tc** or **sr < tr** will remain same because the part for the pattern is same
int inc = C-sc+1 + C * (R-sr); // from DFS of part below source
if (R-sr)%2 == 1
return inc + pattern2(sr, sc)
int pattern2(ll Or, ll Oc) {
/* solve for pattern
.
.
->--^v---<-----<-----<
^----<>------>-------^
->----^v---<-----<---<
^------<>------>-----^
-->-----^xxxxxxxxxxxxx
assuming you enter from bottom left and move along direction suggested by arrows.
Or(Obstacle row) is row of entrance(bottommost row in figure)
Oc(Obstacle column) is column of leftmost 'x'
* /
// a new "round" begins(and previous round ends) when search reaches (r-1,1) from (r, i) for some row r.
if(tc-tr < Oc-Or) // on left side of the diagonal line ending at (Or, Oc)
int rounds = (Or - tr + 1) / 2
// no of "rounds" when it next changes direction.
int inc = (C+1) * 2 * rounds // 2 * (C+1) cells are seen in each round
if Or-tr is odd // approaching "end of round"
if tr == 1 // special handling for case in image 2 above
return inc - C-1 + Oc-Or + tr-tc
return inc - (tc-1)) // distance from "end of round" is tc-1
return inc+tc // passed "end of round"
// on right side of diagonal line
int boundaryRow = Or - Oc + 1 // topmost row beyond which pattern2 disappears
if Or-boundaryRow is odd // case in image 3 above
boundaryRow ++
if tr < boundaryRow // outside boundary, apply pattern1
return pattern1(C, thresh-tr-1, tc-1) + C * (Or-boundaryRow) +Oc - 1
int rounds = (Or-tr-1) / 2
int inc = rounds * 2 * (C + 1)
if Or-tr is odd // going rightwards, away from previous "end of round"
return inc + 1 + tc
return inc + 2 * (C + 1) - tc // going leftwards, towards next "end of round"
}
Apart from this, there are 4 special cases to be handled:
-
sc = 1
The order in which rows below source are traversed does not change. The rows above source are traversed by pattern1, and can be handled as a special case.
-
sr=R
The vertices are traversed in pattern2, and can be handled as a special case.
-
(sr, sc)=(R, C)
The vertices are traversed in pattern1, and can be handled as a special case.
-
R=1 or C=1
Can be handled as a special case.
Solutions:
Tester’s solution is a bit untidy. Editorialist’s solution is based on above ideas.
Setter’s solution
Tester’s Solution
Editorialist’s Solution