Red and Black
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 22 Accepted Submission(s) : 13
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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
原本代码：2014.5.24
1 #include <stdio.h> 2 #include <stdlib.h> 3 #include <string.h> 4 int D,SIGN,Len_X,Len_Y; 5 char Map[100][100]; 6 int Sign[100][100]; 7 8 int Sign_Part(int x,int y) 9 { 10 11 int ii,jj; 12 if(x>=Len_X||x<0||y>=Len_Y||y<0)return 0; 13 if(Map[x][y]=='#')return 0; 14 else Map[x][y]='#'; 15 /*putchar(' '); 16 for(ii=0;ii<Len_X;ii++) 17 { 18 for(jj=0;jj<Len_Y;jj++) 19 printf("%c",Map[ii][jj]); 20 putchar(' '); 21 }*/ 22 Sign[x][y]=1; 23 SIGN++; 24 if(Map[x+1][y]!='#'&&x+1<Len_X&&Sign[x+1][y]!=1) 25 { 26 Sign_Part(x+1,y); 27 Map[x+1][y]='.'; 28 } 29 if(Map[x-1][y]!='#'&&x-1>=0&&Sign[x-1][y]!=1) 30 { 31 Sign_Part(x-1,y); 32 Map[x-1][y]='.'; 33 } 34 if(Map[x][y+1]!='#'&&y+1<Len_Y&&Sign[x][y+1]!=1) 35 { 36 Sign_Part(x,y+1); 37 Map[x][y+1]='.'; 38 } 39 if(Map[x][y-1]!='#'&&y-1>=0&&Sign[x][y-1]!=1) 40 { 41 Sign_Part(x,y-1); 42 Map[x][y-1]='.'; 43 } 44 return 0; 45 46 } 47 48 int main() 49 { 50 int i,j,Begin_x,Begin_y; 51 while(scanf("%d%d",&Len_Y,&Len_X)!=EOF) 52 { 53 if(Len_X==0&&Len_Y==0)break; 54 for(i=0;i<Len_X;i++) 55 { 56 getchar(); 57 for(j=0;j<Len_Y;j++) 58 { 59 scanf("%c",&Map[i][j]); 60 if(Map[i][j]=='@') 61 {Begin_x=i;Begin_y=j;} 62 } 63 } 64 SIGN=0; 65 memset(Sign,0,sizeof(Sign)); 66 Sign_Part(Begin_x,Begin_y); 67 printf("%d ",SIGN); 68 69 } 70 71 return 0; 72 }
修改：2015.5.8
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 using namespace std; 5 int D,SIGN,Len_X,Len_Y; 6 char Map[100][100]; 7 void BFS(int x,int y) 8 { 9 if(Map[x][y]=='#')return ; 10 else 11 { 12 SIGN++; 13 Map[x][y]='#'; 14 BFS(x-1,y); 15 BFS(x+1,y); 16 BFS(x,y-1); 17 BFS(x,y+1); 18 } 19 } 20 21 int main() 22 { 23 int i,j,Begin_x,Begin_y; 24 while(scanf("%d%d",&Len_Y,&Len_X)!=EOF) 25 { 26 if(Len_X==0&&Len_Y==0)break; 27 for(i=0;i<=Len_X+1;i++) 28 { 29 for(j=0;j<=Len_Y+1;j++) 30 { 31 if(i==0||j==0||i==Len_X+1||j==Len_Y+1)Map[i][j]='#'; 32 else 33 { 34 scanf(" %c",&Map[i][j]); 35 if(Map[i][j]=='@') 36 {Begin_x=i;Begin_y=j;} 37 } 38 39 } 40 } 41 SIGN=0; 42 BFS(Begin_x,Begin_y); 43 printf("%d ",SIGN); 44 45 } 46 return 0; 47 }