Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24708 Accepted Submission(s): 14942
Total Submission(s): 24708 Accepted Submission(s): 14942
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.
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)
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
题解：
#include<cstdio>
#include<cstring>
#include<queue>
#include<iostream>
#define INF 99999999
using namespace std;
const int maxn=205;
char s[maxn][maxn];
int vis[maxn][maxn];
int minx=INF;
int n,m;
int dir[4][2]={0,1,-1,0,0,-1,1,0};
int cnt=1;
void dfs(int x,int y){
for(int i=0;i<4;i++){
int tx=x+dir[i][0];
int ty=y+dir[i][1];
if(tx<0||tx>=m||ty<0||ty>=n)
continue;
if(!vis[tx][ty]&&s[tx][ty]!='#'){
cnt++;
vis[tx][ty]=1;
dfs(tx,ty);
}
}
}
int main(){
while(~scanf("%d %d",&n,&m)&&(n+m)){
memset(vis,0,sizeof vis);
cnt=1;
for(int i=0;i<m;i++){
scanf("%s",s[i]);
}
int sx,sy;
int flag=0;
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
if(s[i][j]=='@'){
sx=i,sy=j;
flag=1;
break;
}
}
if(flag) break;
}
vis[sx][sy]=1;
dfs(sx,sy);
printf("%d
",cnt);
}
return 0;
}