Red and Black
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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
4559613
题意:其中'.'表示黑砖;'#'表示红砖;'@'表示当前所在位置。
所求为'@'经过上下左右的移动所能经过的黑色瓷砖的最大个数。
# include<cstdio># include<iostream>using namespace std;int w,h,x,y,cnt,vis[30][30],dx[4]={0,0,1,-1},dy[4]={1,-1,0,0};char c[30][30];void dfs(int X,int Y){ for(int i=0;i<4;i++) { X+=dx[i];Y+=dy[i]; //控制当前位置的移动; if(c[X][Y]=='.'&&0<X&&0<Y&&h>=X&&w>=Y&&!vis[X][Y]) //判断条件; { cnt++; vis[X][Y]=1; //标记为已读(即已经记数,防止重复); dfs(X,Y); //进行下一层的搜索; } X-=dx[i];Y-=dy[i]; //还原初始位置坐标; }}int main(){ while(cin>>w>>h&&w&&h) { int i,j; for(i=1;i<=h;i++) for(j=1;j<=w;j++) { cin>>c[i][j]; //存入数据; vis[i][j]=0; //置零;标记为未访问; if(c[i][j]=='@') { vis[i][j]=1; //标记为已访问; x=i; //记下初始坐标; y=j; } } cnt=1; //因为包含初始位置的黑砖,所以重置为1; dfs(x,y); //进行搜索; cout<<cnt<<endl; } return 0;}
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