Red and Black 【DFS】【连通区域的大小】

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19500    Accepted Submission(s): 11866

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

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

思路 其实就是求连通着的 黑其 有几个

代码

#include<stdio.h>#include<string.h>#include<algorithm>#include<math.h>#include<queue>#include<stack>#include<vector>#define inf 0x3f3f3f#define M 25#define mod 100using namespace std;typedef struct {int x,y,num;}dot;int n,m;int maxnum;int map[M][M];int to[4][2]={0,1,0,-1,1,0,-1,0};void dfs(int x,int y){map[x][y]=0; for(int i=0;i<4;i++){int x1=x+to[i][0];int y1=y+to[i][1];if(map[x1][y1]){maxnum++;dfs(x1,y1); //一开始写成 dfs(x1,y1,sum+1) 这个是求一条路上过程中的最大的个数， }//但是本题的意思是求整个过程的个数 }}int main(){while(scanf("%d%d",&m,&n)&&(n||m)){dot st;memset(map,0,sizeof(map));for(int i=1;i<=n;i++){getchar();for(int j=1;j<=m;j++){char s;scanf("%c",&s);if(s=='.') map[i][j]=1;if(s=='@'){st.x=i;st.y=j; }}}maxnum=1;dfs(st.x,st.y);printf("%d\n",maxnum);}return 0; }