red and black(BFS)

最新推荐文章于 2022-01-12 23:41:22 发布

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最新推荐文章于 2022-01-12 23:41:22 发布

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原文链接：http://www.cnblogs.com/ruruozhenhao/p/8638226.html

本文介绍了一个简单的广搜问题，通过实例讲解了如何在一个由黑色和红色方块组成的矩形房间中，从初始位置出发，计算可以到达的所有黑色方块的数量。提供了一段AC代码，并附带了解题思路。

I - Red and Black

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

Submit Status Practice HDU 1312

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

，

一道简单的广搜题，现在还是对这方面的编程太生疏，其实算法很简单但是自己不熟练所以感觉写的时候有些费力。

AC代码：

 1 #include<bits/stdc++.h>
 2 #define N 25
 3 using namespace std;
 4 int vis[N][N];
 5 char mapn[N][N];
 6 int dir[4][2]={
 7     {1,0},
 8     {-1,0},
 9     {0,1},
10     {0,-1},
11 };
12 int cur=0;
13 int n,m;
14 bool ok(int x,int y)
15 {
16     if(x<0||x>=m||y<0||y>=n||vis[x][y]||mapn[x][y]=='#')
17         return false;
18     return true;
19 }
20 void dfs(int x,int y)
21 {
22     cur++;
23     for(int i=0;i<4;i++)
24     {
25         int fx=x+dir[i][0];
26         int fy=y+dir[i][1];
27         if(ok(fx,fy))
28         {
29             vis[fx][fy]=1;
30             dfs(fx,fy);
31         }
32     }
33 }
34 int main()
35 {
36     //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
37     while(scanf("%d%d",&n,&m)!=EOF&&n&&m)
38     {
39         memset(vis,0,sizeof vis);
40         int x,y;
41         getchar();
42         for(int i=0;i<m;i++)
43         {
44             scanf("%s",mapn[i]);
45             //cout<<mapn[i]<<endl;
46             for(int j=0;j<n;j++)
47             {
48                 if(mapn[i][j]=='@')
49                 {
50                     x=i;
51                     y=j;
52                 }
53                 //cout<<mapn[i][j];
54             }
55             //cout<<endl;
56         }
57         cur=0;
58         vis[x][y]=1;
59         dfs(x,y);
60         printf("%d\n",cur);
61     }
62     return 0;
63 }