POJ-1979&&HDU-1312--Red and Black---DFS深搜

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
45
59
6
13

题意：从@出发，问最多能走多少个’ . ’ （@也算）

简单DFS

#include<iostream>#include <utility>#include<cstring>using namespace std;int n,m;char s[110][110];int num=0;void dfs(int x,int y){    if(x>=0&&x<m&&y>=0&&y<n&&s[x][y]=='.')    {        s[x][y]='#';        num++;        dfs(x,y-1);        dfs(x,y+1);        dfs(x-1,y);        dfs(x+1,y);    }}int main(){    while(cin>>n>>m)    {        if(m==0&&n==0)            break;        num=0;        memset(s,'0',sizeof(s));        int i,j;        int t1=0,t2=0;        for(i=0; i<=m-1; i++)            for(j=0; j<=n-1; j++)            {                cin>>s[i][j];                if(s[i][j]=='@')                {                    t1=i;                    t2=j;                }            }        s[t1][t2]='.';//找到@之后要把它变为'.'        dfs(t1,t2);        cout<<num<<endl;    }}