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…

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…

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…

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,…

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(…

这节课讲动态规划的内容,动态规划是一种通用且有效的算法设计思路,它的主要成分是"子问题"+"重用".它可以用于斐波那契和最短路径等问题的求解上. 一.斐波那契 首先,我们来看下斐波那契问题是什么?传统做法和动态规划法有什么区别? 从上图就能很明显地看出动态规划采用了memorization的思路,将历史计算结果保存下来,这样就能避免递归过程中的重复计算. 我们总结动态规划在求解斐波那契数的内容如下: 记录召回(Memorized calls)只花常数时间.动态规划大…