POJ 1979 Red and Black

Red and Black

Time Limit:1000 MS		Memory Limit:30000 K
Total Submissions:21738		Accepted:11656

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 ctions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which between des 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 shocould 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

Japan 2004 Domestic
Question

It's easy to get started with a deep search question. The question is very simple, that is, you cannot use red bricks. You can search from the start point.

Sample Code

/*****************************************************************************#       COPYRIGHT NOTICE#       Copyright (c) 2014 All rights reserved#       ----Stay Hungry Stay Foolish----##       @author       :Shen#       @name         :POJ 1979#       @file         :G:\My Source Code\¡¾ACM¡¿ÑµÁ·\0630 - ËÑË÷\poj1979.cpp#       @date         :2014/06/30 01:02#       @algorithm    :DFS******************************************************************************/#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long int64;int w, h, cnt;char gird[25][25];//driection u,  d,  r,  l;int dx[4] = { 0,  0,  1, -1};int dy[4] = {-1,  1,  0,  0};inline bool inbound(int l, int r, int x){    return (x >= l && x < r);}inline bool check(int x, int y){    return inbound(0, w, x) && inbound(0, h, y);}void dfs(int x, int y){    if (!check(x, y)) return;    else if (gird[x][y] == '@' || gird[x][y] == '.')    {        cnt++;        gird[x][y] = '#';        for (int i = 0; i < 4; i++)            dfs(x + dx[i], y + dy[i]);    }}void solve(){    memset(gird, 0, sizeof(gird));    int sx = 0, sy = 0; cnt = 0;    for (int i = 0; i < h; i++)    for (int j = 0; j < w; j++)    {        scanf(" %c", &gird[j][i]);        if (gird[j][i] == '@') sx = j, sy = i;    }    dfs(sx, sy);    printf("%d\n", cnt);}int main(){    while (scanf("%d%d", &w, &h))    {        if (w == 0 && h == 0) break;        else solve();    }    return 0;}