HDU 1312 Red and Black

题目链接：HDU 1312

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)

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

题意

输入一个“迷宫”,有红色和黑色方块两种，从给定起点出发，问能到达最多的黑色方块的数量

题解：

DFS水题一道,有点像数鱼塘油井什么的，从起点开始，把周围可走的黑色方块变红：改.为#，同时计数。

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<cstring>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>

using namespace std;

typedef long long ll;

const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 5;
char feld[21][21];
int gx, gy;
int n, m;
int dir[4][2] = { 0,1,1,0,0,-1,-1,0 };
void dfs(int x,int y) {
	int nx, ny;
	for (int i(0); i < 4; i++) {
		nx = x + dir[i][0];
		ny = y + dir[i][1];
		if (nx >= 0 && nx < n&&ny >= 0 && ny < m&&feld[nx][ny] == '.') {
			feld[nx][ny] = '*';
			dfs(nx, ny);
		}
	}
}
int main() {
	while (cin >> m >> n,m) {
		int num = 0;
		memset(feld, 0, sizeof(feld));
		for (int i(0); i < n; i++) {
			cin >> feld[i];
			for (int j(0); j < m; j++)
				if (feld[i][j] == '@')
					gx = i, gy = j;
		}
		feld[gx][gy] = '*';
		dfs(gx,gy);
		for (int i(0); i < n; i++)
			for (int j(0); j < m; j++)
				if (feld[i][j] == '*')
					num++;
		cout << num << endl;
	}
	return 0;
}