一、题目
1、题目描述
2、输入输出
2.1输入
2.2输出
3、原题链接
https://codeforces.com/problemset/problem/196/B
二、解题报告
1、思路分析
考虑如何判断一条路径可以无限走?
我们对朴素的网格dfs改进,改进为可以dfs网格外的区域
如果存在某个 位置 (i % n, j % m) 被访问两次,并且两次的(i, j)不同,则说明进入了一条路径的循环,合法。
2、复杂度
时间复杂度: O(NM)空间复杂度:O(NM)
3、代码详解
#include <bits/stdc++.h>
// #include <ranges>
// #define DEBUG
using i64 = long long;
using u32 = unsigned;
using u64 = unsigned long long;
constexpr int inf32 = 1E9 + 7;
constexpr i64 inf64 = 1E18 + 7;
constexpr double eps = 1e-9;struct DSU {std::vector<int> p;int n;DSU(int _n) : p(_n, -1), n(_n) {}void init () {p.assign(n, -1);}int find(int x) {return p[x] < 0 ? x : p[x] = find(p[x]);}void merge(int x, int y) {int px = find(x), py = find(y);if (px == py) return;if (p[px] > p[py]) std::swap(px, py);p[px] += p[py], p[py] = px;}int size(int x) {return -p[find(x)];}
};constexpr int dir[5] = { -1, 0, 1, 0, -1 };void solve() {int n, m;std::cin >> n >> m;std::vector<std::string> g(n);for (int i = 0; i < n; ++ i) std::cin >> g[i];if (n == 1 && m == 1) {std::cout << "Yes";return;}int stx, sty;for (int i = 0; i < n; ++ i)for (int j = 0; j < m; ++ j) if (g[i][j] == 'S') {stx = i, sty = j;break;}auto pos = [&m](int i, int j) {return i * m + j;};std::vector<std::pair<int, int>> st, vis(n * m, { inf32, inf32 });st.emplace_back(stx, sty);vis[pos(stx, sty)] = { stx, sty };while (st.size()) {auto [x, y] = st.back();st.pop_back();for (int k = 0; k < 4; ++ k) {auto [nx, ny] = std::pair(x + dir[k], y + dir[k + 1]);auto [nnx, nny] = std::pair(((nx % n) + n) % n, ((ny % m) + m) % m);// assert(nnx >= 0 && nnx < n);// assert(nny >= 0 && nny < m);if (g[nnx][nny] != '#') {if (vis[pos(nnx, nny)].first < inf32) {if(vis[pos(nnx, nny)] != std::pair(nx, ny)) {std::cout << "Yes";return;}}else {vis[pos(nnx, nny)] = { nx, ny };st.emplace_back(nx, ny);}}}}std::cout << "No";
}auto FIO = []{std::ios::sync_with_stdio(false);std::cin.tie(nullptr);std::cout.tie(nullptr);return 0;
} ();int main() {#ifdef DEBUGfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);#endif int t = 1;// std::cin >> t;while (t --)solve();return 0;
}