re不是这道题你什么意思啊

#include <iostream>
#include <queue>
#include <cstring>

using namespace std;

const int MAXN = 1005;
const int INF = 1e9;

int n, m;
int map[MAXN][MAXN];
int dx[4] = {0, 1, 0, -1}; // 方向数组，右、下、左、上
int dy[4] = {1, 0, -1, 0};
int dist[MAXN][MAXN][105]; // 距离数组，记录每个位置每个方向的最小拐弯次数

struct State {
int x, y, dir, turns;
};

bool isValid(int x, int y) {
return x >= 0 && x < n && y >= 0 && y < m && map[x][y] == 0;
}

int bfs(int x1, int y1, int x2, int y2) {
queue<State> q;
for (int i = 0; i < 4; ++i) {
dist[x1][y1][i] = 0;
q.push({x1, y1, i, 0});
}

while (!q.empty()) {
    State curr = q.front();
    q.pop();

    if (curr.x == x2 && curr.y == y2) {
        return curr.turns;
    }

    for (int i = 0; i < 4; ++i) {
        int nx = curr.x + dx[i];
        int ny = curr.y + dy[i];
        int nturns = curr.turns + (curr.dir != i ? 1 : 0);

        if (isValid(nx, ny) && dist[nx][ny][i] > nturns) {
            dist[nx][ny][i] = nturns;
            q.push({nx, ny, i, nturns});
        }
    }
}

return INF; // 如果无法到达终点，返回一个很大的数

}

int main() {
cin >> n >> m;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
cin >> map[i][j];
for (int k = 0; k < 4; ++k) {
dist[i][j][k] = INF;
}
}
}

int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;

int result = bfs(x1 - 1, y1 - 1, x2 - 1, y2 - 1);
cout << (result == INF ? -1 : result) << endl;

return 0;

}