Red and Black
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 50913 | Accepted: 27001 |
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)
The end of the input is indicated by a line consisting of two zeros.
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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13 题意：@是搜索起点，#不能走，.可以走，.走过一次后会变为#，问从@开始在棋盘上一共可以走几步（@起点算一步）
#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
using namespace std;
int dir[4][2]={{0,-1},{0,1},{1,0},{-1,0}};
string a[105];
int n,m,cnt;
int check(int x,int y)
{
if(x>=0&&x<n&&y>=0&&y<m&&a[x][y]!=‘#‘)
return 1;
else
return 0;
}
void dfs(int x,int y)
{
if(check(x,y)==0)
return ;
else
{
a[x][y]=‘#‘;
cnt++;
for(int i=0;i<4;i++)
{
int dx,dy;
dx=x+dir[i][0];
dy=y+dir[i][1];
dfs(dx,dy);
}
}
}
int main()
{
while(cin>>m>>n&&n&&m)
{
for(int i=0;i<n;i++)
cin>>a[i];
cnt=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(a[i][j]==‘@‘)
{
dfs(i,j);
break;
}
}
}
cout<<cnt<<endl;
}
return 0;
}
原文地址：https://www.cnblogs.com/-citywall123/p/11290235.html