题意

传送门 LeeCode 1728 猫和老鼠 II

题解

任意图上博弈，可参考 Games on arbitrary graphs。具体而言，将博弈双方位置加之先后手状态看作任意图上的一个节点，并根据状态转移建立反图。对于可以确定胜负态的节点，以其为起点，使用类似拓扑序更新的操作不断递推其余节点的状态。对于$n\times m$的方格，任意图博弈求解时间复杂度为$O((nm{)}^2\max (n,m))$。

#include <bits/stdc++.h>
using namespace std;class Solution {public:bool canMouseWin(vector<string>& grid, int catJump, int mouseJump) {int n = grid.size(), m = grid[0].size();int mx = -1, my = -1;int cx = -1, cy = -1;int fx = -1, fy = -1;for (int i = 0; i < n; ++i) {for (int j = 0; j < m; ++j) {auto c = grid[i][j];if (c == 'M') {mx = i, my = j;}if (c == 'C') {cx = i, cy = j;}if (c == 'F') {fx = i, fy = j;}}}const vector<int> dx = {0, 0, -1, 1};const vector<int> dy = {-1, 1, 0, 0};int pos_num = n * m;int state_num = pos_num * pos_num * 2;vector<vector<int>> g(state_num);auto get = [&](int u, int v, int who) {return u + v * pos_num + who * pos_num * pos_num;};auto judge = [&](int x, int y) {return 0 <= x && x < n && 0 <= y && y < m && grid[x][y] != '#';};auto ed = [&](int cv, int mv) {int fv = fx * m + fy;return cv == mv || cv == fv || mv == fv;};for (int cv = 0; cv < pos_num; ++cv) {for (int mv = 0; mv < pos_num; ++mv) {if (ed(cv, mv)) {continue;}for (int who = 0; who < 2; ++who) {int w = who == 0 ? cv : mv;int sx = w / m, sy = w % m;int lim = who == 0 ? catJump : mouseJump;for (int k = 0; k < (int)dx.size(); ++k) {int x = sx, y = sy;for (int d = 0; d <= lim; ++d) {if (!judge(x, y)) {break;}if (who == 0) {g[get(x * m + y, mv, who ^ 1)].push_back(get(cv, mv, who));} else {g[get(cv, x * m + y, who ^ 1)].push_back(get(cv, mv, who));}x += dx[k];y += dy[k];}}}}}vector<int> indeg(state_num);for (int u = 0; u < state_num; ++u) {for (int v : g[u]) {indeg[v] += 1;}}vector<int> win(state_num, -1);function<void(int)> dfs = [&](int v) {for (int u : g[v]) {if (win[u] == -1) {if (win[v] == 0) {win[u] = 1;} else {indeg[u] -= 1;if (indeg[u] == 0) {win[u] = 0;}}if (win[u] != -1) {dfs(u);}}}};for (int cv = 0; cv < pos_num; ++cv) {int x = cv / m, y = cv % m;if (judge(x, y)) {for (int who = 0; who < 2; ++who) {int v = get(cv, cv, who);win[v] = who ^ 1;dfs(v);}}}for (int v = 0; v < pos_num; ++v) {int x = v / m, y = v % m;int f = fx * m + fy;if (judge(x, y) && v != f) {int u = get(f, v, 1);win[u] = 0;dfs(u);int w = get(v, f, 0);win[w] = 0;dfs(w);}}int res = win[get(cx * m + cy, mx * m + my, 1)];return res == 1;}
};