poj 1979 Red and Black

Red and Black Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 16188 Accepted: 8474

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

Sample Output

45 59 6 13 #include<iostream> using namespace std; char map[22][22]; int w, h; int startx, starty; int sum; int dx[] = {1,-1,0,0}; int dy[] = {0,0,1,-1}; int judge(int a, int b) { if(a<0 || b<0 || a>=h || b>=w) return 0; return 1; } void dfs(int x, int y) { map[x][y] = '#'; int i; for(i=0;i<4;i++) { int ax = x+dx[i]; int ay = y+dy[i]; if(judge(ax, ay) && map[ax][ay]=='.') { sum++; dfs(ax, ay); } } } int main() { int i, j; freopen("e:\\data.txt", "r", stdin); freopen("e:\\out.txt", "w", stdout); while(cin>>w>>h && w && h) { sum = 1; for(i=0;i<h;i++) { for(j=0;j<w;j++) { cin>>map[i][j]; if(map[i][j] == '@') { startx = i; starty = j; } } } dfs(startx, starty); cout<<sum<<endl; } }

 

转载于:https://www.cnblogs.com/w0w0/archive/2012/05/08/2490662.html

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