hdu-1312-Red and Black
2015-08-13 20:37
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Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 13464 Accepted Submission(s): 8338
[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
[align=left]Recommend[/align]
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题目分析:
最简单的dfs 就是让你 从@出发最对能走多少步 可以四个方向走但是遇到#不能走
#include<cstdio> #include<cstring> char a[22][22]; int x,y; int n,m; int cnt; void dfs(int x,int y) { if(a[x][y]=='#') return; if(x>=1&&x<=m&&y>=1&&y<=n) { a[x][y]='#';// 把走过的路都标记成墙不能再走了 cnt++;// 每走一步就加一 dfs(x+1,y);// 向下遍历 dfs(x-1,y);// 向上遍历 dfs(x,y-1);// 向左遍历 dfs(x,y+1);// 向右遍历 就是每个方向都走一遍直到遇到 # 转方向 } } int main() { while(~scanf("%d%d",&n,&m)&&n|m) { int i,j; memset(a,0,sizeof(a)); for(i=1;i<=m;i++) { scanf("%s",a[i]+1);// 记住这里一定要加一因为 你的初始位置是 1 1 如果不加仪的话初始位置就 //是1 0了 因为这里选择的是一行一行的输入每行是一个字符串 字符串的首地址就是 0 所以这里要加1 变成 1 1 for(j=1;j<=n;j++) { if(a[i][j]=='@') { x=i; y=j; } } } cnt=0; dfs(x,y); printf("%d\n",cnt); } return 0; }
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