Red and Black
Time Limit:1000 MS |
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Memory Limit:30000 K |
Total Submissions:21738 |
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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;}