【HDU】1312--Red and Black（DFS）

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Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21053    Accepted Submission(s): 12830

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

 

 

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int w,h;int v[25][25];char str[25][25];int ans=0;int fx[4]={0,0,1,-1};int fy[4]={1,-1,0,0};void dfs(int x,int y){v[x][y]=1;for(int i=0;i<4;i++){int xx=x+fx[i],yy=y+fy[i];if(xx>=0&&xx<h&&yy>=0&&yy<w&&!v[xx][yy]&&str[xx][yy]=='.'){ans++;dfs(xx,yy);}}}int main(){while(~scanf("%d%d%*c",&w,&h),w,h){ans=0;for(int i=0;i<h;i++)scanf("%s",str[i]);memset(v,0,sizeof(v));for(int i=0;i<h;i++)for(int j=0;j<w;j++){if(!v[i][j]&&str[i][j]=='@'){ans++;dfs(i,j);}} printf("%d\n",ans);}return 0;}