Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 21327    Accepted Submission(s): 12986 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 45 59 6 13   模板题 #include<iostream> #include<cmath> #include<cstring> #include<vector> #include<stdlib.h> #include<stdio.h> #include<algorithm> #include <set> #include <list> #include <deque> #include<sstream> #include<time.h> #define pi 3.1415926 #define N 2005 #define M 15 #define INF 0x6f6f6f6f #define lson l , m , rt << 1 #define rson m + 1 , r , rt << 1 | 1 using namespace std; typedef long long ll; const int day =21252; const int mod=1000000007; const int maxn = 55555; int n,m,cnt; bool map[25][25]; int to[4][2]={{1,0},{-1,0},{0,1},{0,-1}}; void dfs(int x,int y) { cnt++; map[x][y]=0; for(int i=0;i<4;i++) { int q,w; q=x+to[i][0]; w=y+to[i][1]; if(q>=0&&w>=0&&q<m&&w<n&&map[q][w]) { dfs(q,w); } } } int main() { while(~scanf("%d %d",&n,&m),n,m) { getchar(); int di,dj; for(int i=0;i<m;i++) { for(int j=0;j<n;j++) { char c; scanf("%c",&c); if(c=='#') map[i][j]=0; else if(c=='.') map[i][j]=1; else { di=i; dj=j; map[i][j]=1; } } getchar(); } cnt=0; dfs(di,dj); printf("%d\n",cnt); } return 0; }