深搜：Red and Black

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本文介绍了一个迷宫寻路问题，通过深度优先搜索算法来计算从起始位置可达的所有黑色方块数量。输入包括迷宫的宽度和高度以及迷宫布局，输出为可达黑色方块的总数。

Red and Black

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 60000/30000K (Java/Other)

Total Submission(s) : 20 Accepted Submission(s) : 17

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)
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

Source

PKU

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#define Max 52

using namespace std;

int data[Max][Max];
int vis[Max][Max];
int a,b,count=0;
int sum=0,tmp=0;
struct mod{
    int e,w,n,s;
}pas[Max][Max];
void sign(){
	int i,j;
    for(i=1;i<=a;i++)
		for(j=1;j<=b;j++){
		    if(!(data[i][j]&1))
				pas[i][j].w=1;
			if(!(data[i][j]&4))
				pas[i][j].e=1;
			if(!(data[i][j]&2))
				pas[i][j].n=1;
			if(!(data[i][j]&8))
				pas[i][j].s=1;
		}
}
void dfs(int x,int y){
    if(!x||!y||x==a+1||y==b+1||vis[x][y])
		return ;
	//int der[4][2] = {{0,1},{0,-1},{-1,0},{1,0}};
    vis[x][y]=1;
	tmp++;
	if(pas[x][y].w)
		dfs(x,y-1);
	if(pas[x][y].e)
		dfs(x,y+1);
	if(pas[x][y].n)
		dfs(x-1,y);
	if(pas[x][y].s)
		dfs(x+1,y);
}
int main(){
	int i,j;
    cin>>a>>b;
	for(i=1;i<=a;i++)
		for(j=1;j<=b;j++)
			scanf("%d",&data[i][j]);
	sign();
    for(i=1;i<=a;i++)
		for(j=1;j<=b;j++){
			//printf("aa");
		    if(vis[i][j])
				continue;
			//printf("bb");
			count++;
			dfs(i,j);
			if(tmp>sum)
				sum=tmp;
			tmp=0;
		}
	cout<<count<<endl<<sum<<endl;
    return 0;
}