UTPC 2021 L Maze Game-526互联

若图中存在点使得删去它后 \(S, T\) 不连通，那么 A 可以一步获胜。

否则，双方都不会删去一个点使得删去它后会产生一个点使得删去它后 \(S, T\) 不连通。那么到最后图上会剩下两条 \(S \to T\) 的不交路径。此时一方无论如何操作都会使得另一方获胜。

因为这是二分图，所以这两条路径的并的点数一定为偶数。那么只用判断初始时非障碍格数量的奇偶性，就能知道到达终态步数的奇偶性。

时间复杂度 \(O(\sum nm)\)。

注意不能直接套割点的板子。

code

// Problem: L - Maze Game
// Contest: AtCoder - UTPC 2021
// URL: https://atcoder.jp/contests/utpc2021/tasks/utpc2021_l
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define ID(x, y) (((x) - 1) * m + (y))
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<int, int> pii;

const int maxn = 110;
const int maxm = 10050;

int n, m, dep[maxm], low[maxm], fa[maxm];
char s[maxn][maxn];
vector<int> G[maxm];

void dfs(int u) {
	low[u] = dep[u];
	for (int v : G[u]) {
		if (!dep[v]) {
			fa[v] = u;
			dep[v] = dep[u] + 1;
			dfs(v);
			low[u] = min(low[u], low[v]);
		} else {
			low[u] = min(low[u], dep[v]);
		}
	}
}

void solve() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n * m; ++i) {
		vector<int>().swap(G[i]);
		dep[i] = low[i] = fa[i] = 0;
	}
	int cnt = 0;
	for (int i = 1; i <= n; ++i) {
		scanf("%s", s[i] + 1);
		for (int j = 1; j <= m; ++j) {
			cnt += (s[i][j] != '#');
		}
	}
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			if (i < n && s[i][j] != '#' && s[i + 1][j] != '#') {
				G[ID(i, j)].pb(ID(i + 1, j));
				G[ID(i + 1, j)].pb(ID(i, j));
			}
			if (j < m && s[i][j] != '#' && s[i][j + 1] != '#') {
				G[ID(i, j)].pb(ID(i, j + 1));
				G[ID(i, j + 1)].pb(ID(i, j));
			}
		}
	}
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			if (s[i][j] == 'S') {
				dep[ID(i, j)] = 1;
				dfs(ID(i, j));
			}
		}
	}
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			if (s[i][j] == 'G') {
				int u = ID(i, j);
				while (fa[fa[u]]) {
					if (low[u] >= dep[u] - 1) {
						puts("Alice");
						return;
					}
					u = fa[u];
				}
			}
		}
	}
	puts((cnt & 1) ? "Alice" : "Bob");
}

int main() {
	int T = 1;
	scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}