最开始很容易陷入贪心的误区（直接去考虑什么情况下走最优），但我们很容易发现，两个玩家的所有状态 $60^4 = 12960000$是在 $1e8$ 极限下的有限状态，故而暴力（ $BFS$ 搜索）即可。

#include <bits/stdc++.h>
#define int long long
using namespace std;

int T = 1;
const int N = 60 + 1;
const int MoveX[] = {-1, 0, 1, 0}, MoveY[] = {0, 1, 0, -1};
int n;
char graph[N][N];

int dis[N][N][N][N];
vector<pair<int, int>> players;
queue<array<int, 4>> q;

void Bfs() {
	int x0 = players[0].first, y0 = players[0].second;
	int x1 = players[1].first, y1 = players[1].second;
	q.push({x0, y0, x1, y1});
	memset(dis, 127, sizeof(dis));
	dis[x0][y0][x1][y1] = 0;
	while (!q.empty()) {
		x0 = q.front()[0], y0 = q.front()[1];
		x1 = q.front()[2], y1 = q.front()[3];
		q.pop();
		if (x0 == x1 && y0 == y1) {
			cout << dis[x0][y0][x1][y1];
			return;
		}
		for (int k = 0; k < 4; k++) {
			int nx0 = x0 + MoveX[k], ny0 = y0 + MoveY[k];
			int nx1 = x1 + MoveX[k], ny1 = y1 + MoveY[k];
			if (nx0 < 1 || nx0 > n) nx0 = x0;
			if (ny0 < 1 || ny0 > n) ny0 = y0;
			if (nx1 < 1 || nx1 > n) nx1 = x1;
			if (ny1 < 1 || ny1 > n) ny1 = y1;
			if (graph[nx0][ny0] == '#') nx0 = x0, ny0 = y0;
			if (graph[nx1][ny1] == '#') nx1 = x1, ny1 = y1;
			if (dis[x0][y0][x1][y1] + 1 < dis[nx0][ny0][nx1][ny1]) {
				dis[nx0][ny0][nx1][ny1] = dis[x0][y0][x1][y1] + 1;
				q.push({nx0, ny0, nx1, ny1});
			}
		}
	}
	cout << -1;
}

void Solve() {
	cin >> n;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			cin >> graph[i][j];
			if (graph[i][j] == 'P') {
				players.push_back({i, j});
				//graph[i][j] = '.';
			}
		}
	}
	Bfs();
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	
	while (T--) {
		Solve();
	}
	return 0;
}