杭电1312 Red and Black【递归搜索】

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

 

 

#include<stdio.h>#include<string.h>int n,m;int hash[30][30];char map[30][30];void DFS(int x,int y){    if(x>n||y>m||hash[x][y]||map[x][y]=='#')//    hash[][]==1 如果已经走过就说明已经循环了         return;    hash[x][y]=1;        DFS(x-1,y);    DFS(x+1,y);    DFS(x,y-1);    DFS(x,y+1);}int main(){    while(scanf("%d%d%*c",&m,&n),n+m)//注意m，n的顺序，且要加%*c否则出错因为后面接下来的是字符输入，如果不加回车将会以字符读入。        {        int x,y,i,j;        memset(hash,0,sizeof(hash));        for(i=0;i<30;i++)        {            for(j=0;j<30;j++)            {                map[i][j]='#';            }        }//注意初始化,为了边界         for(i=1;i<=n;i++)        {            for(j=1;j<=m;j++)            {                scanf("%c",&map[i][j]);                if(map[i][j]=='@')                {                    x=i;                    y=j;                }            }            getchar();            //注意加getchar（）否则会将空行当字符输入。        }        DFS(x,y);        int sum=0;        for(i=1;i<=n;i++)        {            for(j=1;j<=m;j++)            {                if(hash[i][j])                {                    sum++;                }            }        }        printf("%d\n",sum);    }     return 0;}

 

#include<stdio.h>int m,n;int cnt;char a[1010][1001];void dfs(int x,int y){    if(a[x][y]=='#') return ;    if(x > n || x < 1 || y > m || y < 1 )        return ;    cnt++;    a[x][y]='#';    dfs(x-1,y);    dfs(x+1,y);    dfs(x,y+1);    dfs(x,y-1);}int main(){    int i,j;    int x,y;    while(scanf("%d%d",&m,&n),m|n)    {        cnt=0;        for(i=1;i<=n;++i)        {            getchar();            for(j=1;j<=m;++j)            {                scanf("%c",a[i]+j);                if(a[i][j]=='@')                    x=i,y=j;            }        }        dfs(x,y);        printf("%d\n",cnt);    }    return 0;}