Red and Black
| Time Limit: 1000MS |
|
Memory Limit: 30000K |
| 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, 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
45
59
6
13
Source
Japan 2004 Domestic
题解
很简单的一道上手的深搜题目，题目意思很简单，就是不能走红色的砖。从起点开始深搜就行了。
代码示例
/*****************************************************************************
# COPYRIGHT NOTICE
# Copyright (c) 2014 All rights reserved
# ----Stay Hungry Stay Foolish----
#
# @author :Shen
# @name :POJ 1979
# @file :G:\My Source Code\??ACM?????·\0630 - ???÷\poj1979.cpp
# @date :2014/06/30 01:02
# @algorithm :DFS
******************************************************************************/
#include
#include
#include
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; }