POJ 1979 Red and Black （dfs）

Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 24100 Accepted: 13020

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

dfs 裸题

AC代码如下：

////  POJ 1979 Red and Black////  Created by TaoSama on 2015-02-19//  Copyright (c) 2014 TaoSama. All rights reserved.//#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>#define CLR(x,y) memset(x, y, sizeof(x))using namespace std;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int N = 1e5 + 10;int n, m, sx, sy;char a[25][25];bool vis[25][25];int d[4][2] = { -1, 0, 1, 0, 0, 1, 0, -1};void dfs(int x, int y) {vis[x][y] = 1;for(int i = 0; i < 4; ++i) {int nx = x + d[i][0], ny = y + d[i][1];if(!vis[nx][ny] && nx >= 1 && nx <= n && ny >= 1        && ny <= m && a[nx][ny] == '.') dfs(nx, ny);}}int main() {#ifdef LOCALfreopen("in.txt", "r", stdin);//freopen("out.txt","w",stdout);#endifios_base::sync_with_stdio(0);while(cin >> m >> n && n != 0 && m != 0) {memset(vis, 0, sizeof vis);for(int i = 1; i <= n; ++i) {for(int j = 1; j <= m; ++j) {cin >> a[i][j];if(a[i][j] == '@') sx = i, sy = j;}}dfs(sx, sy);int ans = 0;for(int i = 1; i <= n; ++i)for(int j = 1; j <= m; ++j)if(vis[i][j]) ++ans;cout << ans << endl;}return 0;}