import Queue
​class Solution(object):def minimalSteps(self, maze):""":type maze: List[str]:rtype: int"""n = len(maze)m = len(maze[0])p = []#记录所有特殊点for i in range(n):for j in range(m):if maze[i][j] in ['S','T','M','O']:p.append((i,j,maze[i][j]))dxs = [0,1,0,-1]dys = [1,0,-1,0]def bfs(x, y):q = Queue.Queue()q.put((x,y))dis = [[10000 for i in range(m)] for j in range(n)]dis[x][y] = 0while not q.empty():_ = q.get()x = _[0]; y = _[1]for dx, dy in zip(dxs, dys):nx = x + dx; ny = y + dyif nx < 0 or nx == n or ny < 0 or ny == m: continueif maze[nx][ny] == '#': continueif dis[nx][ny] > dis[x][y] + 1:dis[nx][ny] = dis[x][y] + 1q.put((nx,ny))res = []for i, j, _ in p:res.append(dis[i][j])return res#计算特殊点之间的最短距离tag = {}dis = []for idx, (i, j, t) in enumerate(p):dis.append(bfs(i, j))if not t in tag: tag[t] = []tag[t].append(idx)sidx = tag['S'][0]tidx = tag['T'][0]#特殊处理 M 不存在的情况if not 'M' in tag:ans = dis[sidx][tidx]if ans == 10000:ans = -1return ans#计算 S 到所有 M 点之间的最短距离Mnum = len(tag['M'])Onum = len(tag['O'])dp = [[10000 for i in range(Mnum)] for i in range(1<<Mnum)]for i in range(Mnum):midx = tag['M'][i]s = 1 << i for j in range(Onum):oidx = tag['O'][j]dp[s][i] = min(dp[s][i], dis[sidx][oidx] + dis[oidx][midx])# 预处理 M 点之间的最短距离Mdis = [[10000 for i in range(Mnum)] for j in range(Mnum)]for i in range(Mnum):midx1= tag['M'][i]for j in range(Mnum):midx2 = tag['M'][j]for k in range(Onum):oidx = tag['O'][k]Mdis[i][j] = min(Mdis[i][j], dis[midx1][oidx] + dis[oidx][midx2])#状态压缩DPfor s in range(1<<Mnum):for j in range(Mnum):if s & (1<<j) == 0: continuefor k in range(Mnum):if s & (1<<k) != 0: continuens = s | (1<<k)dp[ns][k] = min(dp[ns][k], dp[s][j] + Mdis[j][k])# 统计结果ans = 10000fs = (1<<Mnum) - 1for j in range(Mnum):midx = tag['M'][j]ans = min(ans, dp[fs][j] + dis[midx][tidx])if ans == 10000: ans = -1return ans

class Solution {public:int dx[4] = {1, -1, 0, 0};int dy[4] = {0, 0, 1, -1};int n, m;
​bool inBound(int x, int y) {return x >= 0 && x < n && y >= 0 && y < m;}
​vector<vector<int>> bfs(int x, int y, vector<string>& maze) {vector<vector<int>> ret(n, vector<int>(m, -1));ret[x][y] = 0;queue<pair<int, int>> Q;Q.push({x, y});while (!Q.empty()) {auto p = Q.front();Q.pop();int x = p.first, y = p.second;for (int k = 0; k < 4; k++) {int nx = x + dx[k], ny = y + dy[k];if (inBound(nx, ny) && maze[nx][ny] != '#' && ret[nx][ny] == -1) {ret[nx][ny] = ret[x][y] + 1;Q.push({nx, ny});}}}return ret;}
​int minimalSteps(vector<string>& maze) {n = maze.size(), m = maze[0].size();// 机关 & 石头vector<pair<int, int>> buttons, stones;// 起点 & 终点int sx, sy, tx, ty;for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (maze[i][j] == 'M') {buttons.push_back({i, j});}if (maze[i][j] == 'O') {stones.push_back({i, j});}if (maze[i][j] == 'S') {sx = i, sy = j;}if (maze[i][j] == 'T') {tx = i, ty = j;}}}int nb = buttons.size();int ns = stones.size();vector<vector<int>> start_dist = bfs(sx, sy, maze);
​// 边界情况：没有机关if (nb == 0) {return start_dist[tx][ty];}// 从某个机关到其他机关 / 起点与终点的最短距离。vector<vector<int>> dist(nb, vector<int>(nb + 2, -1));// 中间结果vector<vector<vector<int>>> dd(nb);for (int i = 0; i < nb; i++) {vector<vector<int>> d = bfs(buttons[i].first, buttons[i].second, maze);dd[i] = d;// 从某个点到终点不需要拿石头dist[i][nb + 1] = d[tx][ty];}
​for (int i = 0; i < nb; i++) {int tmp = -1;for (int k = 0; k < ns; k++) {int mid_x = stones[k].first, mid_y = stones[k].second;if (dd[i][mid_x][mid_y] != -1 && start_dist[mid_x][mid_y] != -1) {if (tmp == -1 || tmp > dd[i][mid_x][mid_y] + start_dist[mid_x][mid_y]) {tmp = dd[i][mid_x][mid_y] + start_dist[mid_x][mid_y];}}}dist[i][nb] = tmp;for (int j = i + 1; j < nb; j++) {int mn = -1;for (int k = 0; k < ns; k++) {int mid_x = stones[k].first, mid_y = stones[k].second;if (dd[i][mid_x][mid_y] != -1 && dd[j][mid_x][mid_y] != -1 ) {if (mn == -1 || mn > dd[i][mid_x][mid_y] + dd[j][mid_x][mid_y]) {mn = dd[i][mid_x][mid_y] + dd[j][mid_x][mid_y];}}}dist[i][j] = mn;dist[j][i] = mn;}}
​// 无法达成的情形for (int i = 0; i < nb; i++) {if (dist[i][nb] == -1 || dist[i][nb + 1] == -1) return -1;}// dp 数组， -1 代表没有遍历到vector<vector<int>> dp(1 << nb, vector<int>(nb, -1));for (int i = 0; i < nb; i++) {dp[1 << i][i] = dist[i][nb];}// 由于更新的状态都比未更新的大，所以直接从小到大遍历即可for (int mask = 1; mask < (1 << nb); mask++) {for (int i = 0; i < nb; i++) {// 当前 dp 是合法的if (mask & (1 << i)) {for (int j = 0; j < nb; j++) {// j 不在 mask 里if (!(mask & (1 << j))) {int next = mask | (1 << j);if (dp[next][j] == -1 || dp[next][j] > dp[mask][i] + dist[i][j]) {dp[next][j] = dp[mask][i] + dist[i][j];}}}}}}
​int ret = -1;int final_mask = (1 << nb) - 1;for (int i = 0; i < nb; i++) {if (ret == -1 || ret > dp[final_mask][i] + dist[i][nb + 1]) {ret = dp[final_mask][i] + dist[i][nb + 1];}}
​return ret;}
};