POJ1979-Red and Black

Red and Black
Time Limit: 1000MS  Memory Limit: 30000K
Total Submissions: 22250  Accepted: 12011

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.

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)
The end of the input is indicated by a line consisting of two zeros.

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

Japan 2004 Domestic

 

 

问题描述：有一间长方形的房子，地上铺了红色、黑色两种颜色的正方形瓷砖。你站在其中一块黑色的瓷砖上，只能向相邻的黑色瓷砖移动。请写一个程序，计算你总共能够到达多少块黑色的瓷砖。

算法：设f(x,y)为从点(x,y)出发能够走过的黑瓷砖总数，则
f(x,y) = 1+f(x-1,y)+f(x+1,y)+f(x,y-1)+f(x,y+1)
这里需要注意，凡是走过的瓷砖不能够被重复走过

AC代码：

#include<stdio.h>#include<string.h>//方块四周加白色块，去掉边界判断，//使得递归统一终止于白色块int Rank[22][22];int FindBlack(int c,int l){   if(Rank[c][l]==2)//出口条件(防止重复走)   return 0;   else Rank[c][l]=2;   return 1+FindBlack(c+1,l)+FindBlack(c-1,l)+FindBlack(c,l-1)+FindBlack(c,l+1);//递归函数}int main(){    int i,j,c,l,n,m;    char x;    while(scanf("%d %d",&m,&n),n!=0&&m!=0)    {       max=0;       for(i=0;i<22;i++)       for(j=0;j<22;j++)       Rank[i][j]=2;       for(i=1;i<=n;i++)       {            getchar();          for(j=1;j<=m;j++)          {                scanf("%c",&x);              if(x=='.')              Rank[i][j]=1;              else              {                 if(x=='#')                 Rank[i][j]=2;                 else                 {                     c=i;l=j;                     Rank[i][j]=1;                  }              }          }       }       printf("%d\n",FindBlack(c,l));    }    return 0;}