To solve this problem, we can model it as a variation of the shortest path problem in a grid, where two frogs move simultaneously, and the player wants to minimize the total number of button presses (i.e., moves) such that both frogs reach the destination.
### Approach:
1. **Model the problem as a BFS (Breadth-First Search):**
- Each state in the BFS will represent the positions of both frogs and the total number of moves taken so far.
- The state can be stored as a tuple: `(r1, c1, r2, c2, moves)` where `(r1, c1)` is the position of the first frog and `(r2, c2)` is the position of the second frog.
- The goal is to reach the destination `(R3, C3)` for both frogs.
2. **Movement Rules:**
- Frogs can move in four directions: up, down, left, right.
- If a frog is already at the destination, it can stay there while the other frog continues to move.
- Frogs cannot occupy the same position while moving.
3. **Handling Walls:**
- Frogs cannot move into walls (`#`).
4. **BFS Algorithm:**
- Use a queue to perform BFS.
- Use a visited set to avoid revisiting the same state.
- For each state, generate all possible next states by moving each frog in all valid directions.
- Stop the BFS when both frogs reach the destination.
5. **Edge Cases:**
- If either frog cannot reach the destination, return `-1`.
### Implementation in Python:
```python
from collections import deque
import sys
input = sys.stdin.read
def bfs(grid, N, M, start1, start2, destination):
directions = [(-1, 0), (1, 0), (0, -1), (0, 1)] # up, down, left, right
visited = set()
queue = deque()
# Initial state: (r1, c1, r2, c2, moves)
queue.append((start1[0], start1[1], start2[0], start2[1], 0))
visited.add((start1[0], start1[1], start2[0], start2[1]))
while queue:
r1, c1, r2, c2, moves = queue.popleft()
# Check if both frogs have reached the destination
if (r1, c1) == destination and (r2, c2) == destination:
return moves
# Generate all possible next states
for dr, dc in directions:
nr1, nc1 = r1 + dr, c1 + dc
nr2, nc2 = r2 + dr, c2 + dc
# Check if the new positions are valid for frog 1
if 0 <= nr1 < N and 0 <= nc1 < M and grid[nr1][nc1] != '#':
pos1 = (nr1, nc1)
else:
pos1 = (r1, c1) # Frog stays in place if move is invalid
# Check if the new positions are valid for frog 2
if 0 <= nr2 < N and 0 <= nc2 < M and grid[nr2][nc2] != '#':
pos2 = (nr2, nc2)
else:
pos2 = (r2, c2) # Frog stays in place if move is invalid
# Check if frogs are at the same position (not allowed)
if pos1 == pos2 and pos1 != destination:
continue
# Add the new state to the queue if not visited
new_state = (pos1[0], pos1[1], pos2[0], pos2[1])
if new_state not in visited:
visited.add(new_state)
queue.append((pos1[0], pos1[1], pos2[0], pos2[1], moves + 1))
# If destination is unreachable
return -1
# Main function to read input and process test cases
def main():
data = input().splitlines()
T = int(data[0])
idx = 1
for case in range(1, T + 1):
N, M = map(int, data[idx].split())
idx += 1
R1, C1, R2, C2, R3, C3 = map(int, data[idx].split())
idx += 1
# Adjusting for 0-based indexing
start1 = (R1 - 1, C1 - 1)
start2 = (R2 - 1, C2 - 1)
destination = (R3 - 1, C3 - 1)
grid = []
for _ in range(N):
grid.append(list(data[idx]))
idx += 1
result = bfs(grid, N, M, start1, start2, destination)
print(f"#{case} {result}")
# Run the main function
if __name__ == "__main__":
main()
```
### Explanation of the Code:
- **BFS Function:**
- The `bfs` function performs a breadth-first search to find the shortest number of moves required for both frogs to reach the destination.
- It uses a queue to explore all possible states and a set to track visited states to avoid redundant checks.
- **Main Function:**
- The `main` function reads input, processes each test case, and prints the result for each one.
### Sample Output:
For the given input:
```
1
6 6
6 1 1 6 2 3
....#.
.#..#.
.#..#.
.#..#.
.#.##.
.#....
```
The output would be:
```
#1 13
```
This solution ensures that both frogs reach the destination in the minimum number of moves while avoiding walls and ensuring they do not collide.
该编程解决方案通过构建哈希表，分析给定的Frogs序列（如croak），确保字符遵循特定规律（有k必有a，有a必有o），并处理异常情况，计算最少的青蛙叫声次数。
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