D - Infinite Maze We've got a rectangular n × m-cell maze. Each cell is either passable, or is a wall (impassable). A little boy found the maze and cyclically tiled a plane with it so that the plane became an infinite maze. Now on this plane cell (x,…

197D - Infinite Maze 思路:bfs,如果一个点被搜到第二次,那么就是符合要求的. 用vis[i][j].x,vis[i][j].y表示i,j(i,j是取模过后的值)这个点第一次被搜到的位置,用vis[(next.x%n+n)%n][(next.y%m+m)%m]标记,因为位置可以为负数(绝对值很大的负数). 代码: #include<bits/stdc++.h> using namespace std; ; const int INF=0x3f3f3f3f; char Ma…

We've got a rectangular n × m-cell maze. Each cell is either passable, or is a wall (impassable). A little boy found the maze and cyclically tiled a plane with it so that the plane became an infinite maze. Now on this plane cell (x, y) is a wall if a…

B. Infinite Maze time limit per test  2 seconds memory limit per test  256 megabytes input standard input output  standard output We've got a rectangular n × m-cell maze. Each cell is either passable, or is a wall (impassable). A little boy found the…

Infinite Maze time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output We've got a rectangular n × m-cell maze. Each cell is either passable, or is a wall (impassable). A little boy found the maze…

从起点开始走,对于可以走到的位置,都必定能从这个位置回到起点.这样,对地图进行搜索,当地图中的某一个被访问了两次,就能说明这个地图可以从起点走到无穷远. 搜索的坐标(x,y),x的绝对值可能大于n,的绝对值可能大于m,(x,y)对应在基础地图的位置为rx=(nx%n+n)%n,ry=(ny%m+m)%m We've got a rectangular n × m-cell maze. Each cell is either passable, or is a wall (impassable).…

[题目链接]:http://codeforces.com/problemset/problem/196/B [题意] 给你一个n*m的棋盘; 然后你能够无限复制这个棋盘; 在这个棋盘上你有一个起点s; 然后问你,你能不能从这个起点s开始一直走无限远; [题解] 考虑两个不同棋盘上的点(x1,y1),(x2,y2)(即复制出的另外一个棋盘); 假设他们在第一个棋盘上对应的投影x%n,y%m是相同的; 且它们都能由起点S到达; 即S能到达这两个点; 则可知; 可以从S到达其中的一个点(x1,y1);…

A. Plate Game 如果可以放置一个圆的情况下,先手将圆放置在矩形正中心,那么根据对称性,先手只要放后手的对称的位置即可,也就是先手必胜,否则后手胜. B. Limit 讨论\(n,m\)的大小关系. 虚拟场时当\(n \gt m\)的情况写错了,应该根据\(a_0b_0\)的正负性判断是正无穷还是负无穷. C. Lexicographically Maximum Subsequence 从后往前扫,递增的取. D. Infinite Maze 假设原地图为\(M\),则扩展成\[MM\…

Meandering Through the Maze of MFC Message and Command Routing Paul DiLascia Paul DiLascia is a freelance software consultant specializing in developing C++ applications for Windows. He is the author of Windows++: Writing Reusable Code in C++ (Addiso…

Sept. 10, 2015 Study again the back tracking algorithm using recursive solution, rat in maze, a classical problem. Made a few of mistakes through the practice, one is how to use two dimension array, another one is that "not all return path returns va…