题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1312

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15186    Accepted Submission(s): 9401

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 move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above.

 
Input
The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

'.' - a black tile 
'#' - a red tile 
'@' - a man on a black tile(appears exactly once in a data set) 

 
Output
For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself). 
 
Sample Input
6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0
 
Sample Output
45
59
6
13
 
 #include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
using namespace std;
#define N 25
char mp[N][N];
int go[][] = {{,},{-,},{,-},{,}}; int n,m;
bool ck(int x, int y){
if(x<n&&x>=&&y<m&&y>=) return true;
else return false;
}
bool vis[N][N];
void dfs(int x, int y, int &sum)
{
vis[x][y] = ;
bool fl = ;
for(int i = ; i < ; i++){
int xx = x + go[i][];
int yy = y + go[i][];
if(ck(xx,yy)&&!vis[xx][yy]&&mp[xx][yy]=='.'){
fl = ;
sum = sum+;
dfs(xx,yy,sum);
}
}
if(fl==) return;
}
int main()
{
while(~scanf("%d%d",&m,&n))
{
int x, y;
if(n==&&m==) break;
getchar();
for(int i = ; i < n; i++){
for(int j = ; j < m; j++){
scanf("%c",&mp[i][j]);
if(mp[i][j]=='@'){ x = i; y = j; }
}
getchar();
}
memset(vis,,sizeof(vis));
int sum = ;
dfs(x,y,sum);
printf("%d\n",sum+);
}
return ;
}

Red and Black(dfs水)的更多相关文章

  1. poj1564 Sum It Up dfs水题

    题目描述: Description Given a specified total t and a list of n integers, find all distinct sums using n ...

  2. poj 1979 Red and Black(dfs水题)

    Description There is a rectangular room, covered with square tiles. Each tile is colored either red ...

  3. 【wikioi】1229 数字游戏(dfs+水题)

    http://wikioi.com/problem/1229/ 赤裸裸的水题啊. 一开始我认为不用用完全部的牌,以为爆搜会tle.. 可是我想多了. 将所有状态全部求出,排序后暴力判断即可. (水题有 ...

  4. HDU 1312 Red and Black DFS(深度优先搜索) 和 BFS(广度优先搜索)

    Red and Black Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total ...

  5. HDU1312——Red and Black(DFS)

    Red and Black Problem DescriptionThere is a rectangular room, covered with square tiles. Each tile i ...

  6. 数据结构——HDU1312:Red and Black(DFS)

    题目描述 There is a rectangular room, covered with square tiles. Each tile is colored either red or blac ...

  7. Red and Black(水)

    Red and Black Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Tot ...

  8. HDU 1312 Red and Black (DFS)

    Problem Description There is a rectangular room, covered with square tiles. Each tile is colored eit ...

  9. HDU 1312 Red and Black(DFS,板子题,详解,零基础教你代码实现DFS)

    Red and Black Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) To ...

随机推荐

  1. php-迭代创建级联目录

    方法一代码: path = './a/b/c/d/e/f'; $path_arr = explode('/',$path);//得到数组array('.','a','b','c','d','e','f ...

  2. pwd 命令详解

    pwd 作用: 以绝对路径的方式显示用户当前工作目录,命令将当前目录的全路径名称(从根目录)写入标准输出, 全部目录使用/分隔,第一个/表示根目录, 最后一个/ 表示当前目录. 执行pwd 命令可以立 ...

  3. glibc-commons 依赖解析 版本错误,xxx is duplicate yyy

    glibc-commons 安装了两个版本,导致依赖glibc-commons的很多软件包 被安装了两个版本: 解决办法就是 先清除这些重复的已安装的软件,然后执行 yum update 将 glib ...

  4. Linux 文本编辑器vi命令

    1.Vim Vim  是一个功能强大的全屏幕文本编辑器,是 Linux/UNIX 上最常用的文本编辑器,它的作用是建立.编辑.显示文本文件. Vim 没有菜单,只有命令 2.Vim 工作模式 3.插入 ...

  5. jquery通过ajax查询数据动态添加到select

    function addSelectData() { //select的id为selectId //清空select中的数据 $("#selectId").empty(); $.a ...

  6. php实现MySQL读写分离

    MySQL读写分离有好几种方式 MySQL中间件 MySQL驱动层 代码控制 关于 中间件 和 驱动层的方式这里不做深究  暂且简单介绍下 如何通过PHP代码来控制MySQL读写分离 我们都知道 &q ...

  7. TCP协议(二)——TIME_WAIT状态

    当TCP主动关闭套接字时,采用四步握手机制来彻底关闭连接.如图: 客户端主动关闭连接,发送FIN段到服务端.TCP状态由ESTABLISHED(连接状态)转为FIN_WAIT1(表示,发送的FIN需要 ...

  8. Java中使用LocalDate根据日期来计算年龄

    Java中和日期直接相关的类有很多,平时最常用到的就是java.util package下面的Date和Calendar,需要用到格式的时候还会用到java.text.SimpleDateFormat ...

  9. java多线程(八)-死锁问题和java多线程总结

    为了防止对共享受限资源的争夺,我们可以通过synchronized等方式来加锁,这个时候该线程就处于阻塞状态,设想这样一种情况,线程A等着线程B完成后才能执行,而线程B又等着线程C,而线程C又等着线程 ...

  10. swiper轮播问题之二:默认显示3张图片,中间显示全部两边显示部分

    其二:项目遇到比较有点要求的轮播图,默认显示3张图片,中间显示全部,两边显示部分.如图: 网上找了也没有找到合适的,最后经过自己摸索写了出来,贴出代码分享给大家.         CSS .swipe ...