hdu1312--Red and Black

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

4559613

 

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 | 1using 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;}