Robots on a Grid (Codeforces 1335F)
Tags: Greedy, matrices

There is a rectangular grid of size $$$n \times m$$$. Each cell of the grid is colored black ('0') or white ('1'). The color of the cell $$$(i, j)$$$ is $$$c_{i, j}$$$. You are also given a map of directions: for each cell, there is a direction $$$s_{i, j}$$$ which is one of the four characters 'U', 'R', 'D' and 'L'.  If $$$s_{i, j}$$$ is 'U' then there is a transition from the cell $$$(i, j)$$$ to the cell $$$(i - 1, j)$$$;  if $$$s_{i, j}$$$ is 'R' then there is a transition from the cell $$$(i, j)$$$ to the cell $$$(i, j + 1)$$$;  if $$$s_{i, j}$$$ is 'D' then there is a transition from the cell $$$(i, j)$$$ to the cell $$$(i + 1, j)$$$;  if $$$s_{i, j}$$$ is 'L' then there is a transition from the cell $$$(i, j)$$$ to the cell $$$(i, j - 1)$$$. It is guaranteed that the top row doesn't contain characters 'U', the bottom row doesn't contain characters 'D', the leftmost column doesn't contain characters 'L' and the rightmost column doesn't contain characters 'R'.You want to place some robots in this field (at most one robot in a cell). The following conditions should be satisfied.  Firstly, each robot should move every time (i.e. it cannot skip the move). During one move each robot goes to the adjacent cell depending on the current direction.  Secondly, you have to place robots in such a way that there is no move before which two different robots occupy the same cell (it also means that you cannot place two robots in the same cell). I.e. if the grid is "RL" (one row, two columns, colors does not matter there) then you can place two robots in cells $$$(1, 1)$$$ and $$$(1, 2)$$$, but if the grid is "RLL" then you cannot place robots in cells $$$(1, 1)$$$ and $$$(1, 3)$$$ because during the first second both robots will occupy the cell $$$(1, 2)$$$. The robots make an infinite number of moves.Your task is to place the maximum number of robots to satisfy all the conditions described above and among all such ways, you have to choose one where the number of black cells occupied by robots before all movements is the maximum possible. Note that you can place robots only before all movements.You have to answer $$$t$$$ independent test cases.

Input: The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 5 \cdot 10^4$$$) — the number of test cases. Then $$$t$$$ test cases follow.The first line of the test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 < nm \le 10^6$$$) — the number of rows and the number of columns correspondingly.The next $$$n$$$ lines contain $$$m$$$ characters each, where the $$$j$$$-th character of the $$$i$$$-th line is $$$c_{i, j}$$$ ($$$c_{i, j}$$$ is either '0' if the cell $$$(i, j)$$$ is black or '1' if the cell $$$(i, j)$$$ is white).The next $$$n$$$ lines also contain $$$m$$$ characters each, where the $$$j$$$-th character of the $$$i$$$-th line is $$$s_{i, j}$$$ ($$$s_{i, j}$$$ is 'U', 'R', 'D' or 'L' and describes the direction of the cell $$$(i, j)$$$).It is guaranteed that the sum of the sizes of fields does not exceed $$$10^6$$$ ($$$\sum nm \le 10^6$$$).

Output: For each test case, print two integers — the maximum number of robots you can place to satisfy all the conditions described in the problem statement and the maximum number of black cells occupied by robots before all movements if the number of robots placed is maximized. Note that you can place robots only before all movements.


Sample Inputs:
3
1 2
01
RL
3 3
001
101
110
RLL
DLD
ULL
3 3
000
000
000
RRD
RLD
ULL

Sample Outputs:
2 1
4 3
2 2

Solution (Python):
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = 'x' in file.mode or 'r' not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            (self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr))
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b'\n') + (not b)
            ptr = self.buffer.tell()
            (self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr))
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            (self.buffer.truncate(0), self.buffer.seek(0))

class IOWrapper(IOBase):

    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode('ascii'))
        self.read = lambda : self.buffer.read().decode('ascii')
        self.readline = lambda : self.buffer.readline().decode('ascii')
(sys.stdin, sys.stdout) = (IOWrapper(sys.stdin), IOWrapper(sys.stdout))
input = lambda : sys.stdin.readline().rstrip('\r\n')
from collections import deque

def dfs(alpha, id):
    stack = [alpha]
    cycle = None
    while stack:
        v = stack[-1]
        if not visited[v]:
            visited[v] = True
            u = graph[v]
            if 1:
                if not visited[u]:
                    stack.append(u)
                elif visited[u] and ids[u] != -1:
                    id = ids[u]
                    cycleid = u
                else:
                    cycle = (u, v)
        else:
            break
    if not cycle:
        result = stack
        for i in range(len(result)):
            ids[result[i]] = id
            cyclelength[result[i]] = cyclelength[cycleid]
            cycledistance[result[i]] = (cycledistance[cycleid] - (len(result) - i)) % cyclelength[result[i]]
    else:
        index = []
        result = stack
        for i in range(len(result)):
            ids[result[i]] = id
            if result[i] == cycle[0] or result[i] == cycle[1]:
                index += [i]
        for i in range(len(result)):
            if i < index[0]:
                cycledistance[result[i]] = (i - index[0]) % (index[1] - index[0] + 1)
            else:
                cycledistance[result[i]] = (i - index[0]) % (index[1] - index[0] + 1)
            cyclelength[result[i]] = index[1] - index[0] + 1
for _ in range(int(input())):
    (n, m) = map(int, input().split())
    colors = [-1]
    for i in range(n):
        colors += [int(k) for k in input()]
    graph = [0] * (n * m + 1)
    for i in range(n):
        line = input()
        for j in range(m):
            if line[j] == 'R':
                graph[i * m + j + 1] = i * m + j + 2
            elif line[j] == 'L':
                graph[i * m + j + 1] = i * m + j
            elif line[j] == 'U':
                graph[i * m + j + 1] = (i - 1) * m + j + 1
            else:
                graph[i * m + j + 1] = (i + 1) * m + j + 1
    visited = [False] * len(graph)
    ids = [-1] * len(graph)
    cycledistance = [0] * len(graph)
    cyclelength = [0] * len(graph)
    id = 0
    oo = []
    for i in range(1, len(graph)):
        if not visited[i]:
            dfs(i, id)
            if ids[i] == id:
                id += 1
                oo += [[0] * cyclelength[i]]
    black = 0
    white = 0
    for i in range(1, len(graph)):
        if colors[i] == 0:
            if not oo[ids[i]][cycledistance[i]]:
                black += 1
                oo[ids[i]][cycledistance[i]] = 1
    for i in range(1, len(graph)):
        if colors[i] == 1:
            if not oo[ids[i]][cycledistance[i]]:
                white += 1
                oo[ids[i]][cycledistance[i]] = 1
    print(white + black, black)

Solution (Python):
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = 'x' in file.mode or 'r' not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            (self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr))
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b'\n') + (not b)
            ptr = self.buffer.tell()
            (self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr))
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            (self.buffer.truncate(0), self.buffer.seek(0))

class IOWrapper(IOBase):

    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode('ascii'))
        self.read = lambda : self.buffer.read().decode('ascii')
        self.readline = lambda : self.buffer.readline().decode('ascii')
(sys.stdin, sys.stdout) = (IOWrapper(sys.stdin), IOWrapper(sys.stdout))
input = lambda : sys.stdin.readline().rstrip('\r\n')
from collections import deque

def dfs(alpha, id):
    """"""
    stack = [alpha]
    cycle = None
    while stack:
        v = stack[-1]
        if not visited[v]:
            visited[v] = True
            u = graph[v]
            if 1:
                if not visited[u]:
                    stack.append(u)
                elif visited[u] and ids[u] != -1:
                    id = ids[u]
                    cycleid = u
                else:
                    cycle = (u, v)
        else:
            break
    if not cycle:
        result = stack
        for i in range(len(result)):
            ids[result[i]] = id
            cyclelength[result[i]] = cyclelength[cycleid]
            cycledistance[result[i]] = (cycledistance[cycleid] - (len(result) - i)) % cyclelength[result[i]]
    else:
        index = []
        result = stack
        for i in range(len(result)):
            ids[result[i]] = id
            if result[i] == cycle[0] or result[i] == cycle[1]:
                index += [i]
        for i in range(len(result)):
            if i < index[0]:
                cycledistance[result[i]] = (i - index[0]) % (index[1] - index[0] + 1)
            else:
                cycledistance[result[i]] = (i - index[0]) % (index[1] - index[0] + 1)
            cyclelength[result[i]] = index[1] - index[0] + 1
for _ in range(int(input())):
    (n, m) = map(int, input().split())
    colors = [-1]
    for i in range(n):
        colors += [int(k) for k in input()]
    graph = [0] * (n * m + 1)
    for i in range(n):
        line = input()
        for j in range(m):
            if line[j] == 'R':
                graph[i * m + j + 1] = i * m + j + 2
            elif line[j] == 'L':
                graph[i * m + j + 1] = i * m + j
            elif line[j] == 'U':
                graph[i * m + j + 1] = (i - 1) * m + j + 1
            else:
                graph[i * m + j + 1] = (i + 1) * m + j + 1
    visited = [False] * len(graph)
    ids = [-1] * len(graph)
    cycledistance = [0] * len(graph)
    cyclelength = [0] * len(graph)
    id = 0
    oo = []
    for i in range(1, len(graph)):
        if not visited[i]:
            dfs(i, id)
            if ids[i] == id:
                id += 1
                oo += [[0] * cyclelength[i]]
    black = 0
    white = 0
    for i in range(1, len(graph)):
        if colors[i] == 0:
            if not oo[ids[i]][cycledistance[i]]:
                black += 1
                oo[ids[i]][cycledistance[i]] = 1
    for i in range(1, len(graph)):
        if colors[i] == 1:
            if not oo[ids[i]][cycledistance[i]]:
                white += 1
                oo[ids[i]][cycledistance[i]] = 1
    print(white + black, black)

Solution (Python):
import sys
from array import array

def input():
    return sys.stdin.buffer.readline().decode('utf-8')

def build_scc(V: int, adj: list) -> list:
    (state, low, order) = ([0] * V, [0] * V, [0] * V)
    (group_id, time) = (1, 1)
    groups = []
    for i in range(V):
        if order[i]:
            continue
        (stack, rec) = ([i], [i])
        while rec:
            v = rec[-1]
            if not order[v]:
                low[v] = order[v] = time
                (time, state[v]) = (time + 1, -1)
            elif low[v] > low[adj[v]]:
                low[v] = low[adj[v]]
            for edge_i in range(state[v], 0):
                dest = adj[v]
                if not order[dest]:
                    state[v] = edge_i + 1
                    stack.append(dest)
                    rec.append(dest)
                    break
                elif state[dest] <= 0 and low[v] > order[dest]:
                    low[v] = order[dest]
            else:
                if low[v] == order[v]:
                    g = []
                    while state[v] <= 0:
                        g.append(stack[-1])
                        state[stack.pop()] = group_id
                    group_id += 1
                    if len(g) > 1:
                        groups.append(g)
                rec.pop()
    return groups
t = int(input())
ans = [''] * t
for ti in range(t):
    (n, m) = map(int, input().split())
    N = n * m
    dir_dict = {'L': -1, 'R': 1, 'U': -m, 'D': m}
    (adj, rev) = ([-1] * N, [[] for _ in range(N)])
    color = array('b', [0]) * N
    for i in range(n):
        for (v, c) in enumerate(map(int, input().rstrip()), start=i * m):
            color[v] = c
    for i in range(n):
        for (v, c) in enumerate(input().rstrip(), start=i * m):
            adj[v] = v + dir_dict[c]
            rev[v + dir_dict[c]].append(v)
    sccs = build_scc(N, adj)
    (ans_sum, ans_black) = (0, 0)
    visited = array('b', [0]) * N
    for scc in sccs:
        size = len(scc)
        black = [0] * size
        stack = []
        v = scc[0]
        for i in range(size):
            visited[v] = 1
            stack.append((v, i))
            v = adj[v]
        while stack:
            (v, mod) = stack.pop()
            if color[v] == 0:
                black[mod] = 1
            for dest in rev[v]:
                if visited[dest]:
                    continue
                visited[dest] = 1
                stack.append((dest, mod - 1 if mod > 0 else size - 1))
        ans_sum += size
        ans_black += sum(black)
    ans[ti] = f'{ans_sum} {ans_black}'
sys.stdout.buffer.write('\n'.join(ans).encode('utf-8'))