Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23956 Accepted Submission(s): 14495
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).
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一个人站在一块黑色瓷砖上，他只能前后左右的走，切不可以走到红色瓷砖上。 在一个H*W矩形房间中，当前位置为 @ 点。他只能沿上下左右4个方向走，且只能走 ‘. ’点，不能走‘ # ’点，已走过的 ‘. ’点可重复走，问他最多可走多少 ‘. ’点。 |
#include<string.h>
#include<stdio.h>
#include<algorithm>
#define N 20
int n,m;
char map[N][N];
bool have[N][N];
int count,trueRow,trueCol;
void bfs(int row, int col){
if(row < n && row >= 0 && col < m && col>=0){//越界
if((map[row][col] == '.' || map[row][col] == '@' ) && have[row][col] == false){
have[row][col] = true;//开始的时候没把’@‘加上，盲区了吧
count++; //计数
bfs(row-1,col);//上下左右遍历
bfs(row,col+1);
bfs(row+1,col);
bfs(row,col-1);
}
}
}
int main()
{
int i,j;
scanf("%d%d",&m,&n);
getchar(); //输入n之后的换行符
memset(have,false,sizeof(have));
for(i=0; i<n; i++){
for(j=0; j<m; j++){
scanf("%c", &map[i][j]);
if(map[i][j] == '@'){//确定开始位置
trueRow = i;
trueCol = j;
}
}
getchar();//输入一行之后的换行符
}
count = 0;//全局的能联通的个数
bfs(trueRow, trueCol);//宽度优先遍历
printf("%d\n",count);
return 0;
}