需要自己转化为迷宫问题！
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>using namespace std;
#define MAX1 50005
#define MAX2 350
int dx[] = { 0,0,1,-1 };
int dy[] = { 1,-1,0,0 };
int M, N;
typedef pair<int, int>P;
int maze[MAX2][MAX2];
int d[MAX2][MAX2];
int x[MAX1], y[MAX1], t[MAX1];
int bfs()
{if (maze[0][0] == 0)return -1;//开局就炸queue<P>que;que.push(P(0, 0));d[0][0] = 0;int x, y;while (!que.empty()){P p = que.front(); que.pop();x = p.first; y = p.second;if (maze[x][y] == INT_MAX)return d[x][y]; //安全。for (int i = 0; i < 4; i++){int nx = x + dx[i];int ny = y + dy[i];if (nx >= 0 && ny >= 0 && d[nx][ny] == INT_MAX&&d[x][y] + 1 < maze[nx][ny])//是否可以移动，是否访问过，是否安全。{que.push(P(nx, ny));d[nx][ny] = d[x][y] + 1;}}}return -1;}
int main()
{int n; cin >> n;for (int i = 0; i < n; i++)cin >> x[i] >> y[i] >> t[i];fill(maze[0], maze[0] + MAX2*MAX2, INT_MAX); //构建迷宫，安全位置里放INT_MAXfill(d[0], d[0] + MAX2*MAX2, INT_MAX); //d中放人在该位置的时间，无穷代表没被访问，for (int i = 0; i < n; i++){maze[x[i]][y[i]] = min(maze[x[i]][y[i]], t[i]); //迷宫里放炸的时间，INX_MAX代表安全for (int j = 0; j < 4; j++){int nx = x[i] + dx[j];int ny = y[i] + dy[j];if (nx >= 0 && ny >= 0)maze[nx][ny] = min(maze[nx][ny], t[i]);}}cout << bfs();system("pause");
}