POJ1979 Description There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can mo…

速刷一道DFS Description There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can mo…

Red and Black Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 47466   Accepted: 25523 Description There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a…

题目链接 http://poj.org/problem?id=1979 思路 floodfill问题,使用dfs解决 代码 #include <iostream> #include <cstring> #include <cstdio> using namespace std; const int N = 20; char maze[N][N]; int visit[N][N]; int dir[4][2] = {{-1, 0}, {0 ,1}, {1, 0}, {0,…

Time limit1000 ms Memory limit30000 kB There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on r…

题目在这里 题目意思是这样的,一个人起始位置在    '@'  处,他在途中能到达的地方为 ' .  '     而  '#' 是障碍物,他不能到达. 问途中他所有能到达的   '.'的数量是多少 ??当然,他自己本身也算一个能到达的点. 其中两个样例的结果是这样的走出来的,这是"显而易见"的,哈哈-当然,当图很大的时候,数起来就能费事了. 所用的这个方法叫做FlooFill(洪水覆盖),从它名字来看就是个很暴力直接的方法,只要我能到的地方,我都用水把你淹没了.可以联想一下,在田地里用…

题目链接:http://poj.org/problem?id=1979 深度优先搜索非递归写法 #include <cstdio> #include <stack> using namespace std; , MAX_H = ; ]; int W, H; int DFS(int sx, int sy); int main() { && W != && H != ) { ; i < W; i++) scanf("%s", M…

POJ2386 Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 25366   Accepted: 12778 Description Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <=…

深度优先搜索(depth-first search)是对先序遍历(preorder traversal)的推广.”深度优先搜索“,顾名思义就是尽可能深的搜索一个图.想象你是身处一个迷宫的入口,迷宫中的路每一个拐点有一盏灯是亮着的,你的任务是将所有灯熄灭,按照DFS的做法如下: 1. 熄灭你当前所在的拐点的灯 2. 任选一条路向前(深处)走,每经过一个拐点将灯熄灭直到与之相邻的拐点的灯全部熄灭后,原路返回到某个拐点的相邻拐点灯是亮着的,走到灯亮的拐点,重复执行步骤1 3. 当所有灯熄灭时,结束 将…

深度优先搜索(DFS) 往往利用递归函数实现(隐式地使用栈). 深度优先从最开始的状态出发,遍历所有可以到达的状态.由此可以对所有的状态进行操作,或列举出所有的状态. 1.poj2386 Lake Couting 题意:八连通被认为连接在一起,求总共有多少个水洼? Sample Input: 10 12 W........WW. .WWW.....WWW ....WW...WW. .........WW. .........W.. ..W......W.. .W.W.....WW. W.W.W.…