poj 1979 Red and Black （简单BFS）

Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 28318 Accepted: 15405

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) 
The end of the input is indicated by a line consisting of two zeros. 

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

这题只需要使用简单的BFS进行搜索即可

#include <stdio.h>#include <iostream>#include <string.h>#include <math.h>#include <ctype.h>#include <algorithm>#include <map>#include <math.h>#include <stack>#include <queue>#define Max 25#define inf 100000000using namespace std;char m[Max][Max];int Mov[4][2] = {{0,1},{1,0},{-1,0},{0,-1}};bool vis[Max][Max];int W,H;int dfs(int x,int y){    int res = 1;    vis[x][y] = true;    for (int i=0; i<4; i++) {        int nx = x+Mov[i][0],ny = y + Mov[i][1];        if (nx>=0&&nx<H&&ny>=0&&ny<W) {            if (!vis[nx][ny]&&m[nx][ny] != '#') {                res += dfs(nx, ny);            }        }    }   // printf("%d\n",res);    return res;    //统计个数}int main(){    while (scanf("%d%d",&W,&H)!=EOF&&W+H) {        memset(vis, 0, sizeof(vis));        //getchar();        int xx = -1,yy = -1;        for (int i=0; i<H; i++) {            scanf("%s",m[i]);            for (int j=0; j<W; j++) {                if (m[i][j] == '@') {                    xx = i,yy = j;                }            }        }//        for (int i=0; i<H; i++) {//            printf("%s\n",m[i]);//        }        printf("%d\n",dfs(xx, yy));    }    return 0;}