AcWing 844. 走迷宫
原题链接
简单
走迷宫
#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
typedef struct {
int x;
int y;
} Node;
const int N = 110;
int n, m;
int g[N][N], d[N][N];
int bfs() {
queue<Node> queue;
// 将距离初始化为-1
memset(d, -1, sizeof d);
d[0][0] = 0;
queue.push({0, 0});
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
while (!queue.empty()) {
Node node = queue.front();
queue.pop();
for (int i = 0; i < 4; i++) {
int nex = node.x + dx[i], ney = node.y + dy[i];
// 不越界,并且有路,并且这个格子没有访问过
if (nex >= 0 && nex < n && ney >= 0 && ney < m
&& g[nex][ney] == 0 && d[nex][ney] == -1) {
d[nex][ney] = d[node.x][node.y] + 1;
queue.push({nex, ney});
}
}
}
return d[n - 1][m - 1];
}
/**
* acwing 844-走迷宫
* @return
*/
int main() {
cin >> n >> m;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cin >> g[i][j];
}
}
cout << bfs() << endl;
return 0;
}