HDU 1312 Red and Black（搜索）

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8337    Accepted Submission(s): 5184

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

 

题意：一张地图，给出人所处的位置@，统计此人能到达的块数，人只能朝上下左右走，且只能走黑色砖块即地图上的点，地图上的#代表红色砖块，不能走。最后统计人能到达的砖块数。

#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>const int MAX = 22;char map[MAX][MAX];int res,sx,sy,w,h;void dfs(int x,int y){if(x<0 || y<0 || x>=h || y>=w)return;if(map[x][y]=='#')return;map[x][y] = '#';++res;dfs(x+1,y);dfs(x-1,y);dfs(x,y+1);dfs(x,y-1);}int main(){int i,j;//freopen("in.txt","r",stdin);    while(scanf("%d %d%*c",&w,&h)!=EOF){if(w==0 && h==0)break;for(i=0;i<h;++i){for(j=0;j<w;++j){map[i][j] = getchar();if(map[i][j]=='@'){sx = i;sy = j;}}getchar();}res = 0;dfs(sx,sy);printf("%d\n",res);}    return 0;}