为啥样例不过呢，是第二篇题解的思路

#include <bits/stdc++.h>
using namespace std;

struct node {
    int x, y, time;
} p;
int m, x, y, t, nx, ny, ti[305][305], b[305][305];

int dx[4] = {0, 0, 1, -1}, dy[4] = {1, -1, 0, 0};
queue<node>q;
int main() {
    cin >> m;
    for (int i = 0; i <= 309; i++)
        for (int j = 0; j <= 309; j++)
            ti[i][j] = -1;
    for (int i = 1; i <= m; i++) {
        cin >> x >> y >> t;
        if (t < ti[x][y] || ti[x][y] == -1)
            ti[x][y] = t;
        for (int i = 0; i < 4; i++) {
            nx = x + dx[i];
            ny = y + dy[i];
            if (nx >= 0 && ny >= 0 && (ti[nx][ny] == -1 || t < ti[nx][ny]))
                ti[nx][ny] = t;
        }
    }
    p.x = 0, p.y = 0, p.time = 0, b[0][0] = 1;
    q.push(p);
    while (!q.empty()) {
        p = q.front();
        q.pop();
        for (int i = 0; i < 4; i++) {
            nx = p.x + dx[i], ny = p.y + dy[i];
            if (nx >= 0 && ny >= 0 && b[nx][ny] == 0 && (ti[nx][ny] == -1 || p.time + 1 < ti[nx][ny])) {
                node tx;
                tx.x = nx, tx.y = ny, tx.time = p.time + 1, b[nx][ny] = 1;
                q.push(tx);
                if (ti[nx][ny] == -1) {
                    cout << tx.time << endl;
                    return 0;
                }
            }
        }
    }
    cout << -1 << endl;
    return 0;
}