【POJ 1979 Red and Black dfs 难度:0】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 1321 棋盘问题 dfs 难度:0

http://poj.org/problem?id=1321 注意是在'#'的地方放棋子 矩阵大小不过8*8,即使是8!的时间复杂度也足以承受,可以直接dfs求解 dfs时标注当前点的行和列已被访问,接着搜索行列都未被访问的新点,注意搜索完毕之后标注当前点的行和列未被访问 #include <cstdio> #include <cstring> using namespace std; int n,k; char maz[8][9]; int e[8][8],len[8]; boo…

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 3009 Curling 2.0 回溯,dfs 难度:0

http://poj.org/problem?id=3009 如果目前起点紧挨着终点,可以直接向终点滚(终点不算障碍) #include <cstdio> #include <cstring> using namespace std; ; int maz[maxn][maxn]; int n,m; ] = {,-,,}; ] = {,,,-}; bool in(int x,int y) { && x < n && y >= &&a…

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）

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] = '#';…