hdu1312

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24734    Accepted Submission(s): 14953

Problem 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

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

染色数块问题  bfs板子或者dfs板子都过。。。。

#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mset(a,b) memset(a,b,sizeof(a))
#define sz size()
#define cl clear()
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define PI 3.1415926535897932384626433832795028841971693993751058209749445923078164
typedef long long ll;
typedef pair<int,int> pr;
const int inf = 99999999;
const double eps = 1e-8;
const int dir4[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
const int dir8[8][2] = {{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int maxn = 1;  //易修改 
struct node{
	int x, y;
	node(){
	}
	node(int x,int y)
	{
		this->x = x;
		this->y = y;
	}
//	const operator ? (const & node a) const
//	{
//		return 
//	}
};
int m, n; 
int sx, sy;
char mp[22][22];
int visit[22][22];
int maxS;
int bfs()
{
	queue<node> q;
	q.push(node(sx,sy));
	visit[sx][sy] = 1;
	
	while(!q.empty())
	{
		node t = q.front();
		q.pop();
		
		for(int i = 0;i < 4;i ++)
		{
			int dx = t.x + dir4[i][0];
			int dy = t.y + dir4[i][1];
			
			if(dx < 1 || dy < 1 || dx > n || dy > m || visit[dx][dy] || mp[dx][dy] == '#') continue;
			
			q.push(node(dx,dy));
			visit[dx][dy] = 1;
		}
	}
	int ts = 0;
	for(int i = 1;i <= n;i ++)
	{
		for(int j = 1;j <= m;j ++)
		if(visit[i][j]) ts ++;
	}
	return ts;
}

int main() 
{
	while(cin >> m >> n && m && n)
	{
		maxS = 0;
		mset(visit,0);
		for(int i = 1;i <= n;i ++)
		{
			for(int j = 1;j <= m;j ++)
			{
				cin >> mp[i][j];
				if(mp[i][j] == '@')
				{
					sx = i;
					sy = j;
				}
			}
		}
		
		cout << bfs() << endl;
	}
	return 0;
}