HDU 1312：Red and Black

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17510    Accepted Submission(s): 10642

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

 

一个简单的深搜题目，给出地图，‘.’是可以走的区域，起点是‘@’，问在这个起点开始走可以遇到多少个‘.’;

AC代码：

#include<cstdio>#include<cstring>#include<iostream>#include<math.h>#include<algorithm>using namespace std;char mapp[25][25];bool visit[25][25];int s,w,h;;void DFS(int x,int y){    if(x>=0&&x<h&&y>=0&&y<w&&mapp[x][y]!='#'&&!visit[x][y]) //如果在地图内并且该电是‘.’并且没有遍历过    {        s++;        visit[x][y]=true;                //标记已遍历        DFS(x+1,y);                      //四个方向        DFS(x,y+1);        DFS(x,y-1);        DFS(x-1,y);    }}int main(){    while(cin>>w>>h&&(w||h))    {        int x,y;        s=0;        memset(visit,0,sizeof(visit));  //初始化visit数组        for(int i=0; i<h; i++)        {            cin>>mapp[i];            for(int j=0; j<w; j++)      //对每一行查找起点                if(mapp[i][j]=='@')x=i,y=j;        }        DFS(x,y);                       //深度优先遍历        cout<<s<<endl;    }    return 0;}