HDU1312 Red and Black 【DFS模板】

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

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>using namespace std;/*const int maxn=100;bool vst[maxn][maxn]; // 访问标记int map[maxn][maxn]; // 坐标范围int dir[4][2]= {0,1,0,-1,1,0,-1,0}; // 方向向量，(x,y)周围的四个方向bool CheckEdge(int x,int y) // 边界条件和约束条件的判断{    if(!vst[x][y] && ...) // 满足条件        return 1;    else // 与约束条件冲突        return 0;}void dfs(int x,int y){    vst[x][y]=1; // 标记该节点被访问过    if(map[x][y]==G) // 出现目标态G    {        ...... // 做相应处理        return;    }    for(int i=0; i<4; i++)        ACM Book        BFS 52    {        if(CheckEdge(x+dir[i][0],y+dir[i][1])) // 按照规则生成下一个节点            dfs(x+dir[i][0],y+dir[i][1]);    }    return; // 没有下层搜索节点，回溯}int main(){    ......    return 0;}*/const int maxn=20;char map[maxn][maxn];bool visit[maxn][maxn];int dx[4]= {0,1,0,-1};int dy[4]= {1,0,-1,0};int n, m, sum;void dfs(int x, int y){    int xx, yy;    for(int i=0; i<4; ++i)    {        xx = x + dx[i];        yy = y + dy[i];        if(xx<0 || yy<0 || xx>=n || yy>=m) continue;//越界        if(map[xx][yy] == '.')        {            sum++;            map[xx][yy] = '#'; //表示已访问过，也可以用visit数组标记            dfs(xx, yy);        }    }}int main(){    int fi,fj;    while(~scanf("%d%d",&m,&n) && (n+m))    {        sum=1;        //for(int i=0; i<maxn; ++i)        //  for(int j=0; j<maxn; ++j)        //        map[i][j]=0;        //此处不需初始化，多此一举        for(int i=0; i<n; ++i)        {            for(int j=0; j<m; ++j)            {                cin>>map[i][j];                if(map[i][j]=='@')                {                    fi=i,fj=j;                    map[fi][fj]='#';                }            }            getchar();        }        /*或        for(int i=0;i<n;i++)            scanf("%s",&map[i]);        for(int i=0; i<n; ++i)            for(int j=0; j<m; ++j)                if(map[i][j]=='@')                {                    map[fi][fj]='#';                    dfs(i,j);                }        */        dfs(fi,fj);        cout<<sum<<endl;    }    return 0;}