[AcWing 1112] 迷宫
DFS + 去重
复杂度 \(O(n^{2})\)
点击查看代码
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 100 + 10;
int n;
char g[N][N];
int xa, ya, xb, yb;
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
bool st[N][N];
bool dfs(int x, int y)
{
if (g[x][y] == '#')
return false;
if (x == xb && y == yb)
return true;
st[x][y] = true;
for (int i = 0; i < 4; i ++) {
int a = x + dx[i], b = y + dy[i];
if (a < 0 || a >= n || b < 0 || b >= n || st[a][b])
continue;
if (dfs(a, b))
return true;
}
return false;
}
void solve()
{
cin >> n;
for (int i = 0; i < n; i ++)
for (int j = 0; j < n; j ++)
cin >> g[i][j];
cin >> xa >> ya >> xb >> yb;
memset(st, false, sizeof st);
bool ok = dfs(xa, ya);
cout << (ok ? "YES" : "NO") << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(nullptr);
int T;
cin >> T;
while (T --) {
solve();
}
// solve();
return 0;
}
- 类比 \(BFS\) 走迷宫,\(DFS\) 需要写成递归的形式,用 \(bool\) 类型的 \(st\) 数组用来判断格点是否被搜过,避免重复搜索