水题保平安
Red and Black
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 22640 | Accepted: 12223 |
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
#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <ctype.h>
#include <limits.h>
#include <string.h>
#include <string>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <queue>
#include <stack>
#include <deque>
#include <vector>
#include <set>
//#include <map>
using namespace std;
#define MAXN 20 + 5
char map[MAXN][MAXN];
int w,h;
int vis[MAXN][MAXN];
int num;
void DFS(int x,int y){
if(x-1>=0&&x-1<h&&y>=0&&y<w&&vis[x-1][y]==0){
if(map[x-1][y] == '.'){
num++;
vis[x-1][y] = 1;
DFS(x-1,y);
}
}
if(x>=0&&x<h&&y+1>=0&&y+1<w&&vis[x][y+1]==0){
if(map[x][y+1] == '.'){
num++;
vis[x][y+1] = 1;
DFS(x,y+1);
}
}
if(x+1>=0&&x+1<h&&y>=0&&y<w&&vis[x+1][y]==0){
if(map[x+1][y] == '.'){
num++;
vis[x+1][y] = 1;
DFS(x+1,y);
}
}
if(x>=0&&x<h&&y-1>=0&&y-1<w&&vis[x][y-1]==0){
if(map[x][y-1] == '.'){
num++;
vis[x][y-1] = 1;
DFS(x,y-1);
}
}
}
int main(){
int i,j;
while(~scanf("%d%d",&w,&h)){
if(w==0 && h==0){
break;
}
for(i=0;i<h;i++){
scanf("%s",map[i]);
}
memset(vis,0,sizeof(vis));
int flag = 0;
for(i=0;i<h;i++){
for(j=0;j<w;j++){
if(map[i][j] == '@'){
vis[i][j] = 1;
num = 0;
DFS(i,j);
flag = 1;
break;
}
}
if(flag == 1){
break;
}
}
printf("%d\n",num+1);
}
return 0;
}