Problem 81 » Red and Black

最新推荐文章于 2020-03-19 13:17:11 发布

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最新推荐文章于 2020-03-19 13:17:11 发布

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分类专栏： HDOJ--ACMSteps

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本文介绍了一种基于深度优先搜索的迷宫寻路算法。该算法通过递归地访问迷宫中的每个黑色方块（可达区域），并在访问过的方块上标记避免重复访问，最终统计出从起始位置可达的所有方块数量。输入包括迷宫的宽度和高度及迷宫布局，输出为可达方块总数。

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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

#include<iostream>
using namespace std;
int num,w,h;
char a[21][21];
bool f(int x,int y)
{
   if(x<0||x>=h||y<0||y>=w)
   return 0;
   else
   return 1;
}
void dps(int x,int y)
{
   if(f(x,y)&&a[x][y]=='.')
   {
     a[x][y]='#';
    num++;
    dps(x-1,y);
    dps(x+1,y);
    dps(x,y-1);
    dps(x,y+1);
   }

}
int main()
{
    int i,j;
while(cin>>w>>h&&w!=0&&h!=0)
{
for(i=0;i<h;i++)
   for(j=0;j<w;j++)
   cin>>a[i][j];
num=0;
for(i=0;i<h;i++)
   for(j=0;j<w;j++)
    if(a[i][j]=='@')
    {
      a[i][j]='.';
     dps(i,j);
    }
cout<<num<<endl;
}