Red and Black（DFS）

最新推荐文章于 2020-08-01 10:03:14 发布

木易唐唐

最新推荐文章于 2020-08-01 10:03:14 发布

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分类专栏： DFS(深搜)

本文链接：https://blog.youkuaiyun.com/qq_34457443/article/details/52054225

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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（DFS）解决一个站在黑色方块上的人如何遍历所有可达的黑色方块的问题。该问题在一个由红色和黑色方块组成的矩形房间中进行，目标是计算从起始位置可以达到的所有黑色方块的数量。

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Red and Black

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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

思路：借用DFS（点击打开链接）的知识点解答

#include<cstdio>  
#include<iostream>
#include<string.h>  
#include<algorithm>  
#define M 300  
using namespace std;  
int dx[4]={0,1,-1,0};//x的四个方向 上 下 左 右 
int dy[4]={1,0,0,-1};//y的四个方向 上 下 左 右  
int n, m, x, y, ans;   
bool vis[M][M];  
char map[M][M];  
void F(int xx,int yy)  
{  
    int i,j;  
    for(i=0;i<4;i++)//上下左右4个方向,所以4次  
    {   int xxx=xx+dx[i],  yyy=yy+dy[i];
        if(map[xxx][yyy]=='.' && vis[xxx][yyy]==false )  
         //判断此位置是否为'.'   判断是否来过此位置 
        if( xxx>=0 && yyy>=0 && xxx<m && yyy<n )
{ //判断目前的坐标是否超出范围
            vis[xxx][yyy]=true;  
            ans++;    
            F(xxx,yyy);  
        }  
    }  
}  
int main()  
{  
    while(scanf("%d %d",&n,&m)&&(n||m))  
    {     
        memset(vis,false,sizeof(vis));//把所有的点记为未走过的  
        for(int i=0;i<m;i++)  
        {  
            getchar();  
            for(int j=0;j<n;j++)  
            {  
                scanf("%c",&map[i][j]);  
                if(map[i][j]=='@')//标记@点的坐标  
                {    x=i,y=j;    vis[i][j]=true;    }  
            }  
        }  
        ans=0;  
        F(x,y);  
        cout<<ans+1<<endl;//ans+1是因为@点也在到达范围内  
    }  
    return 0;  
}