Red and Black

最新推荐文章于 2022-01-12 23:41:22 发布

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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。该算法通过递归地探索所有可能路径来寻找从起点到终点的可达黑色方块数量。文章详细解释了如何通过输入地图数据，使用递归函数进行遍历，并记录可达方块的数量。

Red and Black

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

题意：红格子和黑格子。人只能走黑格子，不能越过红色格子，则人能走多少个黑格子。

#include<stdio.h>
#include<string.h>
#define M 300
using namespace std;
int dx[4]={0,-1,1,0};
int dy[4]={-1,0,0,1};
int n,m,x,y,ans;
bool vis[M][M];//标记走过的路
char map[M][M];//输入的字符串
int F(int xx,int yy)
{
    int i,j;
    for(i=0;i<4;i++)
    {
        if(map[xx+dx[i]][yy+dy[i]]=='.'&&vis[xx+dx[i]][yy+dy[i]]==0&&xx+dx[i]>=0&&yy+dy[i]>=0&&xx+dx[i]<m&&yy+dy[i]<n)
        {
            vis[xx+dx[i]][yy+dy[i]]=1;
            ans++;
            F(xx+dx[i],yy+dy[i]);
        }
    }
}
int main()
{
    while(scanf("%d%d",&n,&m)&&(n||m))
    {
        memset(vis,0,sizeof(vis));
        for(int i=0;i<m;i++)
        {
            getchar();
            for(int j=0;j<n;j++)
            {
                scanf("%c",&map[i][j]);
                if(map[i][j]=='@')
                {
                    x=i,y=j;
                    vis[i][j]=1;
                }
            }
        }
        ans=0;
        F(x,y);
        printf("%d\n",ans+1);
    }
    return 0;       
}