POJ1979_Red and Black【DFS】

Red and Black

Time Limit: 1000MS Memory Limit: 30000K

Total Submissions: 23182Accepted: 12504

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

Source

Japan 2004 Domestic

题目大意：给你一个房间，房间由一格格砖组成，有黑砖和红砖，只能站在黑砖上，

现在你站在其中一块黑砖上，且只能向相邻的黑砖上走，问：最多有多少块黑砖可以走

思路：数据规模不大，直接DFS上下左右四个方向

#include<stdio.h>#include<string.h>char map[25][25],vis[25][25];int ans,W,H;void dfs(int i,int j){    if(map[i][j] == '#')        return;    if(i < 0 || i >= H || j < 0 || j >= W)        return;    if(vis[i][j]==0)    {        ans++;        vis[i][j] = 1;        dfs(i-1,j);        dfs(i,j-1);        dfs(i,j+1);        dfs(i+1,j);        return;    }}int main(){    while(~scanf("%d%d",&W,&H) && (W||H))    {        memset(map,0,sizeof(map));        memset(vis,0,sizeof(vis));        for(int i = 0; i < H; i++)        {            getchar();            scanf("%s",map[i]);        }        int posi,posj;        for(int i = 0; i < H; i++)            for(int j = 0; j < W; j++)                if(map[i][j]=='@')                {                    posi = i;                    posj = j;                }        ans = 0;        dfs(posi,posj);        printf("%d\n",ans);    }    return 0;}