Red and Black
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 47795 | Accepted: 25706 |
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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
Source
题目大意：
一个人站在一个黑地板上，然后他可以走四个方向，左下右上，并且他只能走黑色格子，问：他能走到的黑色格子有多少个？
解题思路：
深搜，要加一个标记，标记该地板已经走过了。每次到一个黑色地板ans++。每次都要对地图和标记变量、ans初始化。
吐槽：日本人的思路真奇特，先输入列、再输入行。硬是半天没有反应过来。
//Red and Black
/*
'.' - a black tile
'#' - a red tile
'@' - a man on a black tile(appears exactly once in a data set)
*/
#include<iostream>
#define MAXN 30
using namespace std;
char mapp[MAXN][MAXN];
bool book[MAXN][MAXN]={false};
int nextt[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
int N,M; //N列 M行
int ans=0;
void dfs(int x,int y)
{
for(int i=0;i<4;i++)
{
int gx=x+nextt[i][0];
int gy=y+nextt[i][1];
if(gx>=0&&gx<M&&gy>=0&&gy<N&&mapp[gx][gy]=='.'&&book[gx][gy]==false)
{
ans++;
book[gx][gy]=true;
dfs(gx,gy);
}
}
}
int main()
{
while(cin >> N >> M && N && M)
{
int x,y;
ans=0;
for(int i=0;i<MAXN;i++)
for(int j=0;j<MAXN;j++)
book[i][j]=false,mapp[i][j]='\0';
for(int i=0;i<M;i++)
for(int j=0;j<N;j++)
{
cin >> mapp[i][j];
if(mapp[i][j]=='@')
x=i,y=j;
}
book[x][y]=true;
dfs(x,y);
cout << ans+1 << endl;
}
return 0;
}