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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

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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;}};