【【POJ - 1979 】Red and Black（dfs+染色）】的更多相关文章

POJ 1979 Red and Black dfs 难度:0

http://poj.org/problem?id=1979 #include <cstdio> #include <cstring> using namespace std; const int maxn = 21; bool vis[maxn][maxn]; char maz[maxn][maxn]; int n,m; const int dx[4] = {1,-1,0,0}; const int dy[4] = {0,0,1,-1}; int ans; bool in(int…

poj 1979 Red and Black(dfs)

题目链接:http://poj.org/problem?id=1979 思路分析:使用DFS解决,与迷宫问题相似:迷宫由于搜索方向只往左或右一个方向,往上或下一个方向,不会出现重复搜索: 在该问题中往四个方向搜索,会重复搜索,所以使用vis表来标记访问过的点,避免重复搜索. 代码如下: #include <iostream> using namespace std; ; int vis[MAX_N][MAX_N]; char map[MAX_N][MAX_N]; int red_count,…

POJ 1979 Red and Black （红与黑）

POJ 1979 Red and Black (红与黑) Time Limit: 1000MS    Memory Limit: 30000K Description 题目描述 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…

POJ 1979 Red and Black (zoj 2165) DFS

传送门: poj:http://poj.org/problem?id=1979 zoj:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1165 题目大意: 给你初始坐标,标记为'#'的格子不能走,求你能走的所有格子的个数(能走的为'.',初始坐标用'@'表示) 思路: 一看直接DFS就好了嘛.... 好几天没刷题了,回到家来水一发先~ #include<cstdio> #include<cstring> con…

poj 1979 Red and Black（dfs水题）

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 move only…

POJ 1979 Red and Black【DFS】

标准DFS,统计遍历过程中遇到的黑点个数 #include<cstdio> #include<vector> #include<queue> #include<string> #include<map> #include<iostream> #include<cstring> #include<algorithm> using namespace std; typedef long long LL; const…

POJ 1979 Red and Black （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 move only…

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] = '#';…

poj 1979 Red and Black 题解《挑战程序设计竞赛》

地址 http://poj.org/problem?id=1979 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 mo…

OpenJudge/Poj 1979 Red and Black / OpenJudge 2816 红与黑

1.链接地址: http://bailian.openjudge.cn/practice/1979 http://poj.org/problem?id=1979 2.题目: 总时间限制: 1000ms 内存限制: 65536kB 描述 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.…