原题链接在这里:https://leetcode.com/problems/shortest-path-in-binary-matrix/ 题目: In an N by N square grid, each cell is either empty (0) or blocked (1). A clear path from top-left to bottom-right has length k if and only if it is composed of cells C_1, C_2,…

题目如下: In an N by N square grid, each cell is either empty (0) or blocked (1). A clear path from top-left to bottom-right has length k if and only if it is composed of cells C_1, C_2, ..., C_k such that: Adjacent cells C_i and C_{i+1} are connected 8-…

An undirected, connected graph of N nodes (labeled 0, 1, 2, ..., N-1) is given as graph. graph.length = N, and j != i is in the list graph[i] exactly once, if and only if nodes i and j are connected. Return the length of the shortest path that visits…

Given an integer matrix, find the length of the longest increasing path. From each cell, you can either move to four directions: left, right, up or down. You may NOT move diagonally or move outside of the boundary (i.e. wrap-around is not allowed). E…

原题链接在这里:https://leetcode.com/problems/longest-increasing-path-in-a-matrix/ Given an integer matrix, find the length of the longest increasing path. From each cell, you can either move to four directions: left, right, up or down. You may NOT move diag…

We are given a 2-dimensional grid. "." is an empty cell, "#" is a wall, "@" is the starting point, ("a", "b", ...) are keys, and ("A", "B", ...) are locks. We start at the starting poin…

题目链接:https://leetcode.com/problems/shortest-path-visiting-all-nodes/ 题意:已知一条无向图,问经过所有点的最短路径是多长,边权都为1,每个点可能经过多次. 这道题写的时候想简单了,把它当成树的直径来做了,求出一条最长路径len(len上的点只经过一次),2*(点数-1)-len即为答案,竟然过了,后来看了看讨论区发现这不是正解,而且我也没办法证明,感觉是蒙对的.贴下代码: class Solution { public: voi…

题解 题意 给出一个无向图,求遍历所有点的最小花费 分析 1.BFS,设置dis[status][k]表示遍历的点数状态为status,当前遍历到k的最小花费,一次BFS即可 2.使用DP 代码 //BFS class Solution { public: int dis[1<<12][12]; int shortestPathLength(vector<vector<int>>& graph) { int n=graph.size(); if(n==0) re…

题目 非常简单的BFS 暴搜 struct Node { int x; int y; int k; int ans; Node(){} Node(int x,int y,int k,int ans) { this->x=x; this->y=y; this->k=k; this->ans=ans; } }; class Solution { public: int dir[4][2]={{0,-1},{0,1},{1,0},{-1,0}}; int vis[105][105]; i…

设计最短路径 用bfs 天然带最短路径 每一个状态是 当前的阶段 和已经访问过的节点 下面是正确但是超时的代码 class Solution: def shortestPathLength(self, graph): """ :type graph: List[List[int]] :rtype: int """ N=len(graph) Q=collections.deque([(1 << x, x) for x in range(…