Discription A maze is represented by a tree (an undirected graph, where exactly one way exists between each pair of vertices). In the maze the entrance vertex and the exit vertex are chosen with some probability. The exit from the maze is sought by D…

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…

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…

题意:给出n*m的矩阵,矩阵由'.'和'#'组成,再给出k,表示需要在'.'处加k堵墙,使得剩下的'.'仍然是连通的 先统计出这个矩阵里面总的点数'.'为sum 因为题目说了一定会有一个解,所以找到一个'.',就开始搜,搜到sum-k个连通的点的时候跳出,剩下的没有访问过的点就全部填成墙 #include<iostream> #include<cstdio> #include<cstring> #include <cmath> #include<sta…

[题目链接] http://codeforces.com/problemset/problem/123/E [题目大意] 给出一棵,给出从每个点出发的概率和以每个点为终点的概率,求出每次按照dfs序从起点到达终点的期望. [题解] 首先对于期望计算有X(x,y)=X(x)*X(y),所以对于每次dfs寻路只要求出其起点到终点的期望步数,乘上起点的概率和终点的概率即可.对于一个固定起点和终点的dfs寻路,我们可以发现如果一个点在必要路径上,那么这条路被走过的期望一定为1,如果不在必要路线上,那么走…

A. Maze 题目链接:http://codeforces.com/contest/377/problem/A time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Pavel loves grid mazes. A grid maze is an n × m rectangle maze where each cell is e…

A. Maze 题目连接: http://codeforces.com/contest/377/problem/A Description Pavel loves grid mazes. A grid maze is an n × m rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and h…

http://codeforces.com/problemset/problem/123/E 题目翻译:(翻译来自: http://www.cogs.pw/cogs/problem/problem.php?pid=1734) 一个迷宫是一棵树(即一张无向图,其中任意两点之间仅有一条路径).迷宫的起点和终点都按照某种概率随机选取.人们会在迷宫中用深度优先搜索的方法搜寻终点.如果有许多条可能的路径,会等概率地选取一条.考虑如下伪代码: DFS(x) if x == exit vertex then…

题目链接:http://codeforces.com/problemset/problem/377/A 题解: 有tot个空格(输入时统计),把其中k个空格变为wall,问怎么变才能使得剩下的空格依然为连通的.把问题反过来,其实就是求tot-k的连通图.dfs:在搜索过的空格中做个标记,同时更新连通个数. 代码如下: #include<cstdio>//hdu3183 CodeForces 377A dfs #include<cstring> #include<cmath&g…

Description Ilya is working for the company that constructs robots. Ilya writes programs for entertainment robots, and his current project is "Bob", a new-generation game robot. Ilya's boss wants to know his progress so far. Especially he is int…