1) Abridged problem statement
- You are given an n×m grid of blocks (beauty values 0–9), separated by streets/avenues (n+1 horizontal lines, m+1 vertical lines).
- A pedestrian starts at the north-west corner (top-left corner of the grid), facing east. A path is given as a string of:
  - L: turn left 90 degrees
  - R: turn right 90 degrees
  - M: move forward by one block along the current street/avenue
- When moving one segment (on M), the pedestrian “passes” the block(s) adjacent to that street segment:
  - If the segment is internal, they pass 2 blocks (one on each side).
  - If it lies on the border, they pass only 1 block.
- The first time a block is passed, add its beauty bij to the total. Each subsequent pass of the same block adds floor(bij/2).
- The path never goes outside the map. Compute the final total satisfaction.

Constraints: 1 ≤ n, m ≤ 100; path length ≤ 500; bij are digits 0–9.

2) Detailed editorial
Key idea:
- The pedestrian moves along gridlines connecting intersections. Each move M traverses exactly one street segment.
- That segment is adjacent to one or two blocks. For each adjacent block, you apply:
  - +bij if it is the first time this block is passed,
  - +floor(bij/2) if it has been passed before.
- Track how many times each block was passed (or whether it was passed before) and sum contributions.

Coordinate system and direction:
- Let intersections be addressed by (x, y) where:
  - x in [0, m], y in [0, n]
  - (0, 0) is the north-west (top-left) corner (start position), facing east.
- Let blocks be addressed by 0-based indices grid[row][col] where:
  - row in [0, n-1] (0 = north/top row), col in [0, m-1] (0 = west/left column).
- Use directions encoded as:
  - 0: east  (dx=+1, dy= 0)
  - 1: south (dx= 0, dy=+1)
  - 2: west  (dx=-1, dy= 0)
  - 3: north (dx= 0, dy=-1)
- Turning:
  - L: dir = (dir + 3) % 4
  - R: dir = (dir + 1) % 4

Which blocks are adjacent to a moved segment?
- Suppose we move from (x, y) to (nx, ny) = (x + dx[dir], y + dy[dir]).
- Horizontal move (y constant):
  - Let cx = min(x, nx).
  - Above block (if y > 0):     grid[y-1][cx]
  - Below block (if y < n):     grid[y][cx]
- Vertical move (x constant):
  - Let cy = min(y, ny).
  - Left block  (if x > 0):     grid[cy][x-1]
  - Right block (if x < m):     grid[cy][x]
- The sample C++ code implements the same logic via four direction-specific if-branches.

Algorithm:
- Read n, m, the n grid lines of digits, and the path string.
- Maintain:
  - current intersection (x, y) = (0, 0),
  - current direction dir = 0 (east),
  - visited[n][m] (count or boolean), initialized to 0,
  - total satisfaction = 0.
- For each character in the path:
  - If L or R: change dir accordingly.
  - If M:
    - Compute (nx, ny) = (x + dx[dir], y + dy[dir]).
    - Identify the adjacent block(s) using the cases above.
    - For each adjacent block (r, c):
      - beauty = grid[r][c] (as an integer 0..9).
      - If visited[r][c] == 0: add beauty; else add beauty // 2.
      - visited[r][c] += 1.
    - Set (x, y) = (nx, ny).
- Output total satisfaction.

Correctness:
- Every street segment traversal is counted exactly once, and each adjacent block receives a pass count increment.
- Border segments naturally contribute only one adjacent block due to bounds checks.
- The rules for first vs subsequent passes are enforced via visited.

Complexity:
- Path length ≤ 500, so O(|path|) operations.
- Memory O(nm) for visited.

3) Provided C++ solution with detailed comments explaining each line
```cpp
#include <bits/stdc++.h>     // Pulls in standard C++ library headers in one include

using namespace std;

// Overload for printing pair<T1, T2> to an output stream (unused here, but provided)
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Overload for reading pair<T1, T2> from an input stream (unused here)
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Overload for reading a vector<T> from an input stream (unused here)
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;             // Read each element sequentially
    }
    return in;
};

// Overload for printing a vector<T> to an output stream (unused here)
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';     // Print each element separated by spaces
    }
    return out;
};

// Problem globals
int n, m;                    // Number of block rows (n) and block columns (m)
vector<string> grid;         // Grid of digits as strings (size n, each length m)
string path;                 // Path consisting of 'L', 'R', 'M'

// Read input: n, m, grid lines, and the path
void read() {
    cin >> n >> m;           // Read dimensions
    grid.resize(n);          // Prepare grid with n rows
    for(int i = 0; i < n; i++) {
        cin >> grid[i];      // Read each row as a string of m digits
    }
    cin >> path;             // Read the path string
}

void solve() {
    // We simulate the movement along intersections (x, y), tracking a direction
    // and a per-block "visited" count to decide whether to add full or half beauty.

    int x = 0, y = 0;        // Start at north-west corner (top-left intersection)
    int dir = 0;             // 0:east, 1:south, 2:west, 3:north (clockwise order)
    int satisfaction = 0;    // Total satisfaction to accumulate

    // Direction vectors for dir = 0..3 as above
    int dx[] = {1, 0, -1, 0};
    int dy[] = {0, 1, 0, -1};

    // visited[r][c] = how many times block at row r, col c has been passed
    vector<vector<int>> visited(n, vector<int>(m, 0));

    // Process each instruction in the path
    for(char c: path) {
        if(c == 'L') {
            // Turn left: dir - 1 modulo 4
            dir = (dir + 3) % 4;
        } else if(c == 'R') {
            // Turn right: dir + 1 modulo 4
            dir = (dir + 1) % 4;
        } else {
            // Move forward one block in the current direction
            int nx = x + dx[dir];
            int ny = y + dy[dir];

            // Determine the block(s) adjacent to the traversed segment and add contributions
            if(dir == 0) {
                // Moving east: horizontal segment from (x,y) to (x+1,y)
                // Adjacent blocks: above (y-1, x) and below (y, x), if they exist
                if(y > 0 && x < m) {
                    int beauty = grid[y - 1][x] - '0';   // Convert digit char to int
                    satisfaction +=
                        (visited[y - 1][x] == 0) ? beauty : beauty / 2;
                    visited[y - 1][x]++;                 // Mark block as passed
                }
                if(y < n && x < m) {
                    int beauty = grid[y][x] - '0';
                    satisfaction += (visited[y][x] == 0) ? beauty : beauty / 2;
                    visited[y][x]++;
                }
            } else if(dir == 1) {
                // Moving south: vertical segment from (x,y) to (x,y+1)
                // Adjacent blocks: left (y, x-1) and right (y, x), if they exist
                if(y < n && x > 0) {
                    int beauty = grid[y][x - 1] - '0';
                    satisfaction +=
                        (visited[y][x - 1] == 0) ? beauty : beauty / 2;
                    visited[y][x - 1]++;
                }
                if(y < n && x < m) {
                    int beauty = grid[y][x] - '0';
                    satisfaction += (visited[y][x] == 0) ? beauty : beauty / 2;
                    visited[y][x]++;
                }
            } else if(dir == 2) {
                // Moving west: horizontal segment from (x,y) to (x-1,y)
                // Adjacent blocks: above (y-1, x-1) and below (y, x-1), if they exist
                if(y > 0 && x > 0) {
                    int beauty = grid[y - 1][x - 1] - '0';
                    satisfaction +=
                        (visited[y - 1][x - 1] == 0) ? beauty : beauty / 2;
                    visited[y - 1][x - 1]++;
                }
                if(y < n && x > 0) {
                    int beauty = grid[y][x - 1] - '0';
                    satisfaction +=
                        (visited[y][x - 1] == 0) ? beauty : beauty / 2;
                    visited[y][x - 1]++;
                }
            } else {
                // dir == 3, moving north: vertical segment from (x,y) to (x,y-1)
                // Adjacent blocks: left (y-1, x-1) and right (y-1, x), if they exist
                if(y > 0 && x > 0) {
                    int beauty = grid[y - 1][x - 1] - '0';
                    satisfaction +=
                        (visited[y - 1][x - 1] == 0) ? beauty : beauty / 2;
                    visited[y - 1][x - 1]++;
                }
                if(y > 0 && x < m) {
                    int beauty = grid[y - 1][x] - '0';
                    satisfaction +=
                        (visited[y - 1][x] == 0) ? beauty : beauty / 2;
                    visited[y - 1][x]++;
                }
            }

            // Update current position to the next intersection
            x = nx;
            y = ny;
        }
    }

    // Output the total satisfaction
    cout << satisfaction << '\n';
}

int main() {
    ios_base::sync_with_stdio(false); // Speed up cin/cout
    cin.tie(nullptr);                 // Untie cin from cout for faster IO

    int T = 1;                        // Single test case (kept as a variable for structure)
    // cin >> T;                      // Could be used if multiple test cases were given
    for(int test = 1; test <= T; test++) {
        read();                       // Read input
        // cout << "Case #" << test << ": "; // Not needed
        solve();                      // Solve the single test
    }

    return 0;                         // Exit program
}
```

4) Python solution (with detailed comments)
```python
import sys

def main():
    data = sys.stdin.read().strip().splitlines()
    it = iter(data)

    # Read dimensions
    n_m = next(it).split()
    n = int(n_m[0])
    m = int(n_m[1])

    # Read grid rows as strings of digits
    grid_strs = [next(it).strip() for _ in range(n)]

    # Convert to integer grid (0..9)
    grid = [[ord(ch) - ord('0') for ch in row] for row in grid_strs]

    # Read path
    path = next(it).strip()

    # Directions: 0:east, 1:south, 2:west, 3:north
    dx = [1, 0, -1, 0]
    dy = [0, 1, 0, -1]

    # Start at top-left intersection, facing east
    x, y = 0, 0
    dir = 0

    # visited[r][c] = number of times block (r,c) has been passed
    visited = [[0] * m for _ in range(n)]

    satisfaction = 0

    for c in path:
        if c == 'L':
            dir = (dir + 3) % 4  # turn left
        elif c == 'R':
            dir = (dir + 1) % 4  # turn right
        else:
            # Move one segment forward
            nx = x + dx[dir]
            ny = y + dy[dir]

            if dir == 0:
                # East: adjacent blocks above (y-1, x) and below (y, x)
                if y > 0 and x < m:
                    r, c0 = y - 1, x
                    beauty = grid[r][c0]
                    satisfaction += beauty if visited[r][c0] == 0 else beauty // 2
                    visited[r][c0] += 1
                if y < n and x < m:
                    r, c0 = y, x
                    beauty = grid[r][c0]
                    satisfaction += beauty if visited[r][c0] == 0 else beauty // 2
                    visited[r][c0] += 1
            elif dir == 1:
                # South: adjacent blocks left (y, x-1) and right (y, x)
                if y < n and x > 0:
                    r, c0 = y, x - 1
                    beauty = grid[r][c0]
                    satisfaction += beauty if visited[r][c0] == 0 else beauty // 2
                    visited[r][c0] += 1
                if y < n and x < m:
                    r, c0 = y, x
                    beauty = grid[r][c0]
                    satisfaction += beauty if visited[r][c0] == 0 else beauty // 2
                    visited[r][c0] += 1
            elif dir == 2:
                # West: adjacent blocks above (y-1, x-1) and below (y, x-1)
                if y > 0 and x > 0:
                    r, c0 = y - 1, x - 1
                    beauty = grid[r][c0]
                    satisfaction += beauty if visited[r][c0] == 0 else beauty // 2
                    visited[r][c0] += 1
                if y < n and x > 0:
                    r, c0 = y, x - 1
                    beauty = grid[r][c0]
                    satisfaction += beauty if visited[r][c0] == 0 else beauty // 2
                    visited[r][c0] += 1
            else:
                # North: adjacent blocks left (y-1, x-1) and right (y-1, x)
                if y > 0 and x > 0:
                    r, c0 = y - 1, x - 1
                    beauty = grid[r][c0]
                    satisfaction += beauty if visited[r][c0] == 0 else beauty // 2
                    visited[r][c0] += 1
                if y > 0 and x < m:
                    r, c0 = y - 1, x
                    beauty = grid[r][c0]
                    satisfaction += beauty if visited[r][c0] == 0 else beauty // 2
                    visited[r][c0] += 1

            # Advance to the next intersection
            x, y = nx, ny

    print(satisfaction)

if __name__ == "__main__":
    main()
```

5) Compressed editorial
- Model intersections as (x, y), blocks as grid[row][col]. Start at (0,0) facing east.
- On each M, move to (x+dx[dir], y+dy[dir]) and score the adjacent block(s) of that segment:
  - Horizontal: use row y−1 (above) and y (below), col = min(x, nx).
  - Vertical: use col x−1 (left) and x (right), row = min(y, ny).
- For each adjacent block, if first time: add bij; else add floor(bij/2). Track with visited[n][m].
- Complexity O(|path|), with straightforward boundary checks for border segments.