题解 | 走迷宫
走迷宫
https://www.nowcoder.com/practice/e88b41dc6e764b2893bc4221777ffe64
#include <iostream>
#include <vector>
#include <queue>
#include <string>
using namespace std;
struct Point {
int x, y;
};
int main() {
int n, m;
cin >> n >> m;
int sx, sy, tx, ty;
cin >> sx >> sy >> tx >> ty;
// 转换为0-indexed
sx--; sy--; tx--; ty--;
vector<string> grid(n);
for (int i = 0; i < n; i++) {
cin >> grid[i];
}
// 距离数组,初始化为-1表示未访问
vector<vector<int>> dist(n, vector<int>(m, -1));
queue<Point> q;
dist[sx][sy] = 0;
q.push({sx, sy});
// 四个方向:上、下、左、右
int dx[4] = {-1, 1, 0, 0};
int dy[4] = {0, 0, -1, 1};
while (!q.empty()) {
Point p = q.front();
q.pop();
// 到达终点,提前结束BFS
if (p.x == tx && p.y == ty) {
break;
}
for (int i = 0; i < 4; i++) {
int nx = p.x + dx[i];
int ny = p.y + dy[i];
if (nx >= 0 && nx < n && ny >= 0 && ny < m && grid[nx][ny] == '.' && dist[nx][ny] == -1) {
dist[nx][ny] = dist[p.x][p.y] + 1;
q.push({nx, ny});
}
}
}
cout << dist[tx][ty] << endl;
return 0;
}
// 64 位输出请用 printf("%lld")
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