HDU 1312 Red and Black【递归】
2015-08-03 19:35
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Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 13111 Accepted Submission(s): 8127
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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)
[align=left]Output[/align]
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).
[align=left]Sample Input[/align]
6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0
[align=left]Sample Output[/align]
45 59 6 13
[align=left]Source[/align]
Asia 2004, Ehime (Japan), Japan
Domestic
题意:
实际就是让求@上下左右方向上的小黑点(。)的个数!
思路:
我们可以用到递归的方法来计算,每次都去找@现在的位置的前后左右方向上的小黑点,如果有则count++,最终得到的结果,就是我们的所求!
代码:
/* 因为需要每次都得找当前位置周围的小黑点,所以我们要将 大问题化成小问题,我们只求目前的@位置的上下左右小黑点的个数 ,这个比较容易,然后每次都是这样,用到的方法都相同,并且有终止条件, 所以我们比较容易想到递归,这个题就用到了递归的方法!!!!! */ #include <stdio.h> char a[22][22]; int n,m,count; void f(int x,int y)//找@所处的位置的小黑点的个数,并进行递归统计 { if(x<1||x>m||y<1||y>n) return; if(a[x][y]=='#') return; count++; a[x][y]='#'; f(x-1,y);//左 f(x+1,y);//右 f(x,y-1);//下 f(x,y+1);//上 } int main() { int i,j,k,x,y; while(scanf("%d%d",&n,&m)&&(n||m)) { count=0; for(i=1;i<=m;i++)//输入字符 { getchar();//吸收回车 for(j=1;j<=n;j++) { scanf("%c",&a[i][j]); if(a[i][j]=='@')//找@ x=i,y=j; } } f(x,y);//计数小黑点 printf("%d\n",count);//输出小黑点的总数 } return 0; }
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