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hdu 1312 Red and Black(深搜)
Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10286 Accepted Submission(s): 6426
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.
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)
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
Source
Asia 2004, Ehime (Japan), Japan Domestic
做好细节工作,迫在眉睫!!!本来测试数据三组都对了,最后一组怎么都不对,调试了一中午,才发现是map数组忘了初始化了。。。。
代码如下:
#include<stdio.h> #include<string.h> char map[1001][1001]; int visit[1001][1001]; int count,n,m; void dfs(int a,int b) { if(a<0||b<0||a>=1001||b>=1001) return ; if(!visit[a-1][b]&&map[a-1][b]=='.') { visit[a-1][b]=1; count++; dfs(a-1,b); // visit[n-1][m]=0; } if(!visit[a][b-1]&&map[a][b-1]=='.') { visit[a][b-1]=1; count++; dfs(a,b-1); // visit[n][m-1]=0; } if(!visit[a+1][b]&&map[a+1][b]=='.') { visit[a+1][b]=1; count++; dfs(a+1,b); // visit[n+1][m]=0; } if(!visit[a][b+1]&&map[a][b+1]=='.') { visit[a][b+1]=1; count++; dfs(a,b+1); // visit[n][m+1]=0; } else return ; } int main() { int i,j; while(~scanf("%d%d",&n,&m),n||m) { count=1; memset(visit,0,sizeof(visit)); memset(map,'#',sizeof(map));//这个一定要加上,就因为这个疏忽,我足足调试了一中午。。。 for(i=0;i<m;++i) { scanf("%s",map[i]); } for(i=0;i<m;++i) { for(j=0;j<n;++j) { if(map[i][j]=='@') { dfs(i,j); printf("%d\n",count); break; } } } } return 0; }
hdu 1312 Red and Black(深搜)
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