red and black(BFS)
I - Red and Black
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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
,
一道简单的广搜题,现在还是对这方面的编程太生疏,其实算法很简单但是自己不熟练所以感觉写的时候有些费力。
AC代码:
1 #include<bits/stdc++.h> 2 #define N 25 3 using namespace std; 4 int vis[N][N]; 5 char mapn[N][N]; 6 int dir[4][2]={ 7 {1,0}, 8 {-1,0}, 9 {0,1}, 10 {0,-1}, 11 }; 12 int cur=0; 13 int n,m; 14 bool ok(int x,int y) 15 { 16 if(x<0||x>=m||y<0||y>=n||vis[x][y]||mapn[x][y]=='#') 17 return false; 18 return true; 19 } 20 void dfs(int x,int y) 21 { 22 cur++; 23 for(int i=0;i<4;i++) 24 { 25 int fx=x+dir[i][0]; 26 int fy=y+dir[i][1]; 27 if(ok(fx,fy)) 28 { 29 vis[fx][fy]=1; 30 dfs(fx,fy); 31 } 32 } 33 } 34 int main() 35 { 36 //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin); 37 while(scanf("%d%d",&n,&m)!=EOF&&n&&m) 38 { 39 memset(vis,0,sizeof vis); 40 int x,y; 41 getchar(); 42 for(int i=0;i<m;i++) 43 { 44 scanf("%s",mapn[i]); 45 //cout<<mapn[i]<<endl; 46 for(int j=0;j<n;j++) 47 { 48 if(mapn[i][j]=='@') 49 { 50 x=i; 51 y=j; 52 } 53 //cout<<mapn[i][j]; 54 } 55 //cout<<endl; 56 } 57 cur=0; 58 vis[x][y]=1; 59 dfs(x,y); 60 printf("%d\n",cur); 61 } 62 return 0; 63 }
永远渴望,大智若愚(stay hungry, stay foolish)