杭电1312 Red and Black（搜索入门）

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9744    Accepted Submission(s): 6066

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
/*搜索入门题*/#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char map[25][25];bool vis[25][25];int s,e;int N,M;int move[4][2]={0,1,1,0,0,-1,-1,0};bool Judge(int x,int y){if(x>=0&&x<N&&y>=0&&y<M)return true;return false;}int step;void DFS(int x,int y){int xx=x;int yy=y;for(int i=0;i<4;i++){x=xx+move[i][0];y=yy+move[i][1];if(Judge(x,y)&&map[x][y]=='.'&&vis[x][y]==0){step++;vis[x][y]=1;DFS(x,y);}}return;}int main(){while(scanf("%d%d",&M,&N)!=EOF){if(M==0&&N==0)break;memset(vis,0,sizeof(vis));memset(map,0,sizeof(map));for(int i=0;i<N;i++){scanf("%s",map[i]);}int x,y;for(int i=0;i<N;i++){for(int j=0;j<M;j++){if(map[i][j]=='@'){x=i;y=j;}}}//printf("%d %d\n",x,y);step=1;vis[x][y]=1;DFS(x,y);printf("%d\n",step);}return 0;}