hdoj-1312-Red and Black(深搜)

最新推荐文章于 2020-10-12 13:34:03 发布

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本文探讨了一个深搜算法的应用实例，用于计算在特定房间布局中，从初始位置出发，仅能通过黑色瓷砖并避开红色瓷砖所能到达的黑色瓷砖数量。详细介绍了输入输出格式、解题思路及代码实现。

Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 12377 Accepted Submission(s): 7698

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

题目大意：

房间由黑色和红色地板铺成，一个人站在黑色地板上，位置为@，他可以向着上下左右四个方向移动，黑色可走，红色不可走，看其可以坐多少个黑色地板。

解题思路：

这是一道图形深搜，深搜也就是递归的一种，其边界条件就是图形的边界还有不能是‘#’，然后在递归调用本身。如何控制上下左右移动呢？那就是建立两个数组，一个存放X的变化，一个存放Y的变化，其值对应上下左右移动X或Y的变化，记住要一一对应。

每深搜递归调用一次，sum++,然后将其标记为#，避免重复。

#include<stdio.h>
int xx[5]={0,0,-1,1};
int yy[5]={1,-1,0,0};//上下左右移动 
char map[22][22];
int n,m;
int sum;
int judge(int x,int y)
{
	if(x<1||x>m||y<1||y>n)
	    return 0;
	if(map[x][y]=='#')
	    return 0;
	else
	    return 1;
}
void DFS(int x,int y)
{
	int i,j;
	int nowx,nowy;
	sum++;
	map[x][y]='#';//标记，避免重复 
	for(i=0;i<4;i++)
	{
		nowx=x+xx[i];
		nowy=y+yy[i];
		if(judge(nowx,nowy))
		    DFS(nowx,nowy);
	}
}
int main()
{
	int stax,stay;
	int i,j;
	while(scanf("%d%d",&n,&m),n+m)
	{
		for(i=1;i<=m;i++)
		{
			getchar();//将换行排除 
		    for(j=1;j<=n;j++)
		    {
		    	scanf("%c",&map[i][j]);
		    	if(map[i][j]=='@')
		    	{
		    		stax=i;
		    		stay=j;
		    	}
		    }
		}
		sum=0;
		DFS(stax,stay);
		printf("%d\n",sum);
	}
	return 0;
}