非常标准的BFS 第一次写错了很多 1.到达4时设置为墙就好了  避免了死循环 2.不用开d数组   在结构体里面就行了 3.结构体初始化函数的写法: Node(int x=0,int y=0,int oil=0):x(x),y(y),oil(oil){} 4.bfs的FOR里面的判断条件可以写的很清晰!就判断可以的  不可以的直接不处理! #include<bits/stdc++.h> using namespace std; int sx,sy,ex,ey; ][],a[][],d2[][…

Nightmare    hdu1072 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8871    Accepted Submission(s): 4270 Problem Description Ignatius had a nightmare last night. He found himself in a labyrinth…

Nightmare Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 4   Accepted Submission(s) : 3 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Ignatius had a nightmare last…

Nightmare Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5647    Accepted Submission(s): 2808 Problem Description Ignatius had a nightmare last night. He found himself in a labyrinth with a ti…

Nightmare Problem Description Ignatius had a nightmare last night. He found himself in a labyrinth with a time bomb on him. The labyrinth has an exit, Ignatius should get out of the labyrinth before the bomb explodes. The initial exploding time of th…

题目链接:Nightmare 题意: 给出一张n*m的图,0代表墙,1代表可以走,2代表起始点,3代表终点,4代表炸弹重置点 问是否能从起点到达终点 分析: 一道很好的DFS题目,炸弹重置点必然最多走一次,可能会陷入无限递归,故有一个重要的剪枝,见代码, 此题采用记忆化搜索(不懂の),注意代码的设计思路 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; ][],step…

传送门 题意 给出一张n*m的图 0.墙 1.可走之路 2.起始点 3.终点 4.时间重置点 问是否能到达终点 分析 我的训练专题第一题,一开始我设个vis数组记录,然后写炸,不能处理重置点根vis的关系,然后看了iaccepted这篇blog,换了一种思路,加了一个很好的剪枝,0ms过,足可显示我的搜索思想差.实现差,切记!切记! trick 1.dfs不能保证最先到达终点的为最短距离,不能打标记 2.剪枝说明:如果当前到达该点时间少于该点剩余时间并且步数大于该点步数,则返回 代码 #incl…

Description Ignatius had a nightmare last night. He found himself in a labyrinth with a time bomb on him. The labyrinth has an exit, Ignatius should get out of the labyrinth before the bomb explodes. The initial exploding time of the bomb is set to 6…

今天将介绍一款自动化测试套件名叫nightmare,他是一个基于phantomjs的测试框架,一个基于phantomjs之上为测试应用封装的一套high level API.其API以goto, refresh, click, type…等简单的常用e2e测试动作封装,使得其语义清晰,简洁.其官方在http://www.nightmarejs.org/. 如果你的项目测试不需要想需求和测试人员理解,那么基于nightmare测试或许是一个好的选择,你的降低测试代码的成本,以及测试套件的部署.我们…

Navigation Nightmare   Description Farmer John's pastoral neighborhood has N farms (2 <= N <= 40,000), usually numbered/labeled 1..N. A series of M (1 <= M < 40,000) vertical and horizontal roads each of varying lengths (1 <= length <= 1…