Poj 1979 Hdu 1312 Red and Black【dfs】

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 13060    Accepted Submission(s): 8097

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

这个题意是，小数点代表可以走的路，# 代表不能走的路，@代表这个人当前位置，给出现在的场地布置，求出这个人最大的活动区间（可以包含当前位置）....

今天刚学过递归，这个算是很简单的 dfs 问题，只要用简单的递归就可以处理完成了.....具体见代码注释......

#include<stdio.h>int t,n,m,i,j,sum;char x[25][25],y;void rab(int a,int b)//主要运行的函数{    if(x[a][b]=='#')//遇到墙，本次递归结束    {        return;    }    if(a<0||a>m-1||b<0||b>n-1)//超出边界，递归结束    {        return;    }    ++sum;//统计    x[a][b]='#';//当前已经统计过的，标记成 墙     rab(a+1,b);//用递归对上下左右四个方向进行搜索，判断，运算.....    rab(a-1,b);//    rab(a,b+1);//    rab(a,b-1);//}int main(){    int a,b;    while(scanf("%d%d",&n,&m),m||n)    {        for(i=0;i<m;++i)        {            getchar();            for(j=0;j<n;++j)            {                scanf("%c",&y);                x[i][j]=y;                if(y=='@')//就一个位置，单独标记上....                {                    a=i;b=j;                }            }        }        sum=0;        rab(a,b);//调用函数计算结果        printf("%d\n",sum);//输出....    }    return 0; } 

/*2016年3月2日21:35 Poj */#include<stdio.h>#include<string.h>int n,m,ans,dx[4]={-1,1,0,0},dy[4]={0,0,1,-1};char map[25][25];/*void dfs(int a,int b){++ans;map[a][b]='#';for(int i=0;i<4;++i){int tx=a+dx[i],ty=b+dy[i];if(tx<0||tx>=n||ty<0||ty>=m||map[tx][ty]=='#'){continue;}dfs(tx,ty);}}*/void dfs(int a,int b){if(a<0||a>=n||b<0||b>=m||map[a][b]=='#'){return;}++ans;map[a][b]='#';for(int i=0;i<4;++i){dfs(a+dx[i],b+dy[i]);}}int main(){//freopen("shuju.txt","r",stdin);while(scanf("%d%d",&m,&n),n|m){memset(map,0,sizeof(map));int x,y;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]=='@'){x=i;y=j;}}}ans=0;dfs(x,y);printf("%d\n",ans);}return 0;}