这道题目难就难在如何想到用最短路径来做
主要是这个题目不能用bfs来写，因为距离并不是1
狄克斯特拉算法很久没写了，有些地方生疏了
且这个题目需要记录三个信息，得用tuple

题目地址

    int dx[] = {0,0,1,-1};
    int dy[] = {1,-1,0,0};
class Solution {
public:
    int minTimeToReach(vector<vector<int>>& moveTime) {
        int n = moveTime.size(); int m = moveTime[0].size();
        vector<vector<int>> dis (n,vector<int> (m,0x3f3f3f3f));
        priority_queue<tuple<int,int,int>,vector<tuple<int,int,int>>,greater<>> q;
        q.emplace(0,0,0);
        while(q.size()){
            auto [d,i,j] = q.top(); q.pop();
            if(i==n-1 && j == m-1){
                return d;
            }
            if(dis[i][j]!=0x3f3f3f3f) continue;
            for(int k=0;k<4;k++){
                int x = i+dx[k], y = j + dy[k];
                if(x<0 || x >= n || y <0 || y>=m) continue;
                int now = max(d,moveTime[x][y]) + 1;
                if(now<dis[x][y]){
                    dis[x][y] = now;
                    q.emplace(now,x,y);
                }
            }
        }
        return -1;
    }

};