POJ 1979：Red and Black

Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 26058 Accepted: 14139

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

4559613

给一张图，@为起始点，'.'能走，‘#’不能走，问一共能走到多少'.'。

在深夜能做到这种水题也真是很令人高兴的事情。

深度搜索水题。

代码：

#include <iostream>#include <algorithm>#include <cmath>#include <vector>#include <string>#include <cstring>#pragma warning(disable:4996)using namespace std;int row,col,sum;char value[30][30];int flag[30][30];void dfs(int x,int y){flag[x][y]=1;if(x>1&&flag[x-1][y]==0&&value[x-1][y]=='.'){dfs(x-1,y);}if(y>1&&flag[x][y-1]==0&&value[x][y-1]=='.'){dfs(x,y-1);}if(x<row&&flag[x+1][y]==0&&value[x+1][y]=='.'){dfs(x+1,y);}if(y<col&&flag[x][y+1]==0&&value[x][y+1]=='.'){dfs(x,y+1);}}void solve1(){int i,j;for(i=1;i<=row;i++){for(j=1;j<=col;j++){if(value[i][j]=='@'){dfs(i,j);return;}}}}void solve2(){int i,j;for(i=1;i<=row;i++){for(j=1;j<=col;j++){if(flag[i][j]){sum++;}}}}int main(){int i,j;while(cin>>col>>row){if(col+row==0)break;memset(flag,0,sizeof(flag));sum=0;for(i=1;i<=row;i++){cin>>value[i]+1;}solve1();solve2();cout<<sum<<endl;}return 0;}