hdu1312 Red and Black

题目链接

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

---

分析：

搜索模板题。BFS和DFS应该都可以，不过BFS比较快。

程序说明：

int nxt[][]：方向数组。

queue<pair<int,int> > que：BFS的队列，用于扩展点。其中，pair的第一维为x，第二维为y。（也就是坐标）

bool book[]：标记数组，避免搜索中重复扩展。

code:

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
int ans=1,n,m,startx,starty,nxt[4][2]={{0,-1},{-1,0},{0,1},{1,0}};
char a[30][30];
bool book[30][30];
void bfs()
{
	queue<pair<int,int> > que; 
	que.push(make_pair(startx,starty));
	book[startx][starty]=true;
	while(!que.empty())
	{
		for(int k=0;k<4;k++)
		{
			int tx=que.front().first+nxt[k][0];
			int ty=que.front().second+nxt[k][1];
			if(tx<0||tx>=n||ty<0||ty>=m) continue;
			if(book[tx][ty]==false&&a[tx][ty]=='.')
			{
				book[tx][ty]=true;
				ans++;
				que.push(make_pair(tx,ty));
			}
		}
		que.pop();
	}
}
int main()
{
	while(~scanf("%d%d",&m,&n)&&(m||n))
	{
		memset(a,0,sizeof(a));
		memset(book,false,sizeof(book));
		ans=1;
		startx=starty=0;
		int flag=0;
		for(int i=0;i<n;i++) 
		{
			getchar();
			for(int j=0;j<m;j++) 
			{
				a[i][j]=getchar();
				if(a[i][j]=='@') startx=i,starty=j;
			} 
		}
		bfs();
		printf("%d\n",ans);
	}
	return 0;
}