Red and Black(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] = '#';…

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…

poj-1979 && hdoj - 1312 Red and Black (简单dfs)

http://poj.org/problem?id=1979 基础搜索. #include <iostream> #include <cstdio> #include <cmath> #include <vector> #include <cstring> #include <string> #include <algorithm> #include <string> #include <set>…

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 >…

题解报告：hdu 1312 Red and Black（简单dfs）

Problem 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…