HDU 1312 Red and Black

Red and Black

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 19   Accepted Submission(s) : 15

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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). 

Sample Input

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

Sample Output

45
59
6
13

#include<iostream>#include<stdio.h>#include<cmath>#include<string.h>#include<string>#include<queue>#include<stack>#include<vector>#include<climits>#include<stdlib.h>using namespace std;const int maxn=30;int i,j,k,n,m;char point[maxn][maxn];int sum,sx,sy;int dir[4][2]={{-1,0},{1,0},{0,1},{0,-1}};void dfs(int x,int y){if(point[x][y]=='#')return;if(x<0||y<0||x>=n||y>=m)return;sum++;point[x][y]='#';for(int i=0;i<4;i++){dfs(x+dir[i][0],y+dir[i][1]);}return;}int main(){while(cin>>m>>n,n&&m){for(int i=0;i<n;i++)for(int j=0;j<m;j++){cin>>point[i][j];if(point[i][j]=='@') sx=i,sy=j;}sum=0;//cout<<flag[1][1];dfs(sx,sy);cout<<sum<<endl;}return 0;}