poj2386（简单dfs）

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Red and Black(简单dfs)

Red and Black Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12519 Accepted Submission(s): 7753 Problem Description There is a rectangular room, covered with square tiles. Each tile is color…

POJ 1979 Red and Black (简单dfs)

题目: 简单dfs,没什么好说的代码: #include <iostream> using namespace std; typedef long long ll; #define INF 2147483647 int w,h; ][]; ][] = {-,,,,,-,,}; ; void dfs(int x,int y){ || x >= h || y < || y >= w || a[x][y] == '#') return; ans++; a[x][y] = '#';…

POJ1573(Robot Motion)--简单模拟+简单dfs

题目在这里题意 : 问你按照图中所给的提示走,多少步能走出来??? 其实只要根据这个提示走下去就行了.模拟每一步就OK,因为下一步的操作和上一步一样,所以简单dfs.如果出现loop状态,只要记忆每个所到的点的第一次的步数,最后总步数减掉它即可 /************************************************************************* > File Name: poj1573.cpp > Author: YeGuoSheng >…

就是求图中有多少个水洼.对图进行dfs遍历,并把是水洼的地方全部标记.然后从下一个是水哇的地方再进行dfs. #include <cstdio> #include <iostream> #include <cstring> using namespace std; ][]; int sum; ][]; , -, -, , , , , }; , , , -, , -, ,}; int row, col; bool check(int x, int y) { &&a…

poj2386 Lake Counting（简单DFS）

转载请注明出处:viewmode=contents">http://blog.csdn.net/u012860063?viewmode=contents 题目链接:http://poj.org/problem?id=1562 --------------------------------------------------------------------------------------------------------------------------------------…

poj2386(简单的dfs/bfs）

题目链接:http://poj.org/problem?id=2386 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 <= N <= 100; 1 <= M <= 100) squares. Each square contains either wa…

POJ1979 Red and Black (简单DFS)

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…