Red and Black
Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12117Accepted Submission(s): 7543
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
4559613
Source
Asia 2004, Ehime (Japan), Japan Domestic
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AC代码：
#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>using namespace std;int tu[4][2]={{1,0},{0,1},{-1,0},{0,-1}};char dfs[22][22];int t1,t2,n,m,sum;void fun(int x,int y){sum++;dfs[x][y]='#';for(int i=0;i<4;++i){int dx=x+tu[i][0];int dy=y+tu[i][1];if(dx<n&dy<m&&dx>=0&&dy>=0&&dfs[dx][dy]=='.'){fun(dx,dy);}}return ;}int main(){int i,j;char c;while(cin>>m>>n&&(n|m)){for(i=0;i<n;++i){for(j=0;j<m;++j){cin>>dfs[i][j];if(dfs[i][j]=='@'){t1=i;t2=j;}}}sum=0;fun(t1,t2);cout<<sum<<'\12';}return 0;}
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，如果困难是堵砖墙，拍拍它说你还不够高。
