HDU1312——Red and Black(DFS)

Red and Black

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
45
59
6
13

题目大意：

　　　　一个瓦片地图，'.'代表黑色的瓦片，'#'代表红色的瓦片，'#'是主人公站的位置，主人公只会下上左右四种移动方式，且只能去黑色的瓦片（初始位置也是黑色的瓦片）；

　　　　求可以去的黑色瓦片个数，包括初始位置。

解题思路：

　　　　简单的DFS，搜索一下与初始位置上下左右相连的所有黑色瓦片，并记录输出即可。

Code；

 1 #include<iostream>
 2 #include<string>
 3 #include<cstdio>
 4 #define MAXN 50
 5 using namespace std;
 6 bool vis[MAXN+10][MAXN+10],is_black[MAXN+10][MAXN+10]; //黑色瓦片标记
 7 char tile[MAXN+10][MAXN+10];
 8 int n,m;
 9 int dfs(int i,int j)
10 {
11 
12     if (is_black[i][j]==0||vis[i][j]==1) return 0; //搜索时遇到已经搜索过的或者红色瓦片则返回，不记录瓦片数。
13     vis[i][j]=1;
14     int sum=1;
15     if (i-1>=1) sum+=dfs(i-1,j); //搜索上下左右四种情况
16     if (i+1<=n) sum+=dfs(i+1,j);
17     if (j-1>=1) sum+=dfs(i,j-1);
18     if (j+1<=m) sum+=dfs(i,j+1);
19     return sum;
20 }
21 int main()
22 {
23     int first_i,first_j;
24     while (cin>>m>>n)
25     {
26         if (m==0&&n==0) break;
27         memset(is_black,0,sizeof(is_black));
28         memset(vis,0,sizeof(vis));
29         getchar();
30         for (int i=1; i<=n; i++)
31         {
32             for (int j=1; j<=m; j++)
33             {
34                 cin>>tile[i][j];
35                 if (tile[i][j]=='@') first_i=i,first_j=j,tile[i][j]='.'; //记录初始位置用于调用DFS，并用题意将初始位置转换成黑色瓦片（貌似没有必要--！）
36                 if (tile[i][j]=='#') is_black[i][j]=0;//用Is_Black数组标记瓦片颜色
37                 else is_black[i][j]=1;
38             }
39             getchar();
40         }
41 
42         printf("%d\n",dfs(first_i,first_j));
43     }
44     return 0;
45 }

 

posted @ 2014-07-13 10:27  Enumz  阅读(314)  评论(0)  编辑  收藏  举报