Submission details
Task:Monikulmio
Sender:NicholasAhman
Submission time:2025-11-19 13:46:57 +0200
Language:Python3 (PyPy3)
Status:READY
Result:100
Feedback
groupverdictscore
#1ACCEPTED100
Test results
testverdicttimescore
#1ACCEPTED0.04 s10details
#2ACCEPTED0.05 s10details
#3ACCEPTED0.06 s10details
#4ACCEPTED0.06 s10details
#5ACCEPTED0.06 s10details
#6ACCEPTED0.05 s10details
#7ACCEPTED0.08 s10details
#8ACCEPTED0.08 s10details
#9ACCEPTED0.08 s10details
#10ACCEPTED0.26 s10details

Code

import sys

def solve():
    # Read all input from stdin
    input_data = sys.stdin.read().split()
    if not input_data:
        return

    iterator = iter(input_data)
    
    try:
        n = int(next(iterator))
        m = int(next(iterator))
        k = int(next(iterator))
        
        points = []
        for _ in range(k):
            # Read y (row) and x (col). 
            # Convert 1-based input to 0-based implementation coordinates.
            y = int(next(iterator)) - 1
            x = int(next(iterator)) - 1
            points.append((y, x))
            
    except StopIteration:
        return

    # Initialize grid with '.' (outside)
    grid = [['.' for _ in range(m)] for _ in range(n)]
    
    # --- PHASE 1: DRAW BOUNDARIES ---
    
    # Helper to get direction sign
    def get_sign(val):
        if val > 0: return 1
        if val < 0: return -1
        return 0

    # Draw edges
    for i in range(k):
        p1 = points[i]
        p2 = points[(i + 1) % k] # Wrap around to start
        
        r1, c1 = p1
        r2, c2 = p2
        
        # Determine steps and direction
        dr = get_sign(r2 - r1)
        dc = get_sign(c2 - c1)
        
        # Determine character based on slope
        char = ''
        if r1 == r2:
            char = '='
        elif c1 == c2:
            char = '|'
        else:
            # Diagonals: 
            # If dr and dc have same sign (both pos or both neg) -> \
            # If different signs -> /
            # Note: Row index increases downwards.
            # (dr=1, dc=1) is down-right (\)
            # (dr=-1, dc=1) is up-right (/)
            if dr == dc:
                char = '\\'
            else:
                char = '/'
        
        # Draw the line segment
        # We draw from p1 up to (but not including) p2 to avoid overwriting
        # the next edge's start char logic, though standardizing vertices later fixes this.
        curr_r, curr_c = r1, c1
        steps = max(abs(r2 - r1), abs(c2 - c1))
        
        for _ in range(steps):
            grid[curr_r][curr_c] = char
            curr_r += dr
            curr_c += dc
            
    # Draw corners (Always '*')
    # Done after edges to ensure corners take precedence
    for r, c in points:
        grid[r][c] = '*'

    # --- PHASE 2: FILL INTERIOR ---
    
    boundary_chars = {'*', '-', '=', '|', '/', '\\'}
    
    for r in range(n):
        for c in range(m):
            # If currently on a boundary, skip logic
            if grid[r][c] in boundary_chars:
                continue
            
            intersections = 0
            
            # Check ray intersection against all polygon edges
            for i in range(k):
                p1 = points[i]
                p2 = points[(i + 1) % k]
                
                y1, x1 = p1
                y2, x2 = p2
                
                # Ray Casting logic:
                # Check if the edge spans across the current row 'r' vertically.
                # We use strict inequality on one side to handle vertices (half-open interval).
                # (y1 > r) != (y2 > r) is true if r is strictly between y1 and y2
                # or equal to the lower one (depending on convention), effectively
                # preventing double counting at vertices.
                
                if (y1 > r) != (y2 > r):
                    # Calculate the x-coordinate where the edge intersects row r
                    # Formula derived from line equation: 
                    # slope = (x2 - x1) / (y2 - y1)
                    # x_inters = x1 + (r - y1) * slope
                    
                    x_inters = x1 + (r - y1) * (x2 - x1) / (y2 - y1)
                    
                    # We only care if the intersection is to the RIGHT of our point
                    if c < x_inters:
                        intersections += 1
            
            # If odd intersections, point is inside
            if intersections % 2 == 1:
                grid[r][c] = '#'

    # --- OUTPUT ---
    for row in grid:
        print("".join(row))

if __name__ == "__main__":
    solve()

Test details

Test 1 (public)

Verdict: ACCEPTED

input
8 9 5
5 2
2 5
5 8
7 8
...

correct output
.........
....*....
.../#\...
../###\..
.*#####*.
...

user output
.........
....*....
.../#\...
../###\..
.*#####*.
...

Test 2 (public)

Verdict: ACCEPTED

input
20 40 4
5 10
5 30
15 30
15 10

correct output
.................................

user output
.................................

Test 3 (public)

Verdict: ACCEPTED

input
20 40 29
8 7
13 2
14 2
9 7
...

correct output
.................................

user output
.................................

Test 4 (public)

Verdict: ACCEPTED

input
20 40 14
5 12
5 25
8 28
13 28
...

correct output
.................................

user output
.................................

Test 5 (public)

Verdict: ACCEPTED

input
20 40 12
3 20
7 16
7 9
11 13
...

correct output
.................................

user output
.................................

Test 6 (public)

Verdict: ACCEPTED

input
9 35 33
2 3
2 8
4 8
4 5
...

correct output
.................................

user output
.................................

Test 7 (public)

Verdict: ACCEPTED

input
30 100 69
6 10
6 14
7 14
7 18
...

correct output
.................................

user output
.................................

Test 8 (public)

Verdict: ACCEPTED

input
40 60 192
11 3
11 5
10 6
11 7
...

correct output
.................................

user output
.................................

Test 9 (public)

Verdict: ACCEPTED

input
50 100 142
1 1
1 7
1 11
1 14
...

correct output
*=====*===*==*...................

user output
*=====*===*==*...................

Test 10 (public)

Verdict: ACCEPTED

input
100 100 1000
10 1
4 7
1 4
1 9
...

correct output
...*====*........................

user output
...*====*........................