• ZOJ Problem Set


    http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1165

    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)


    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

    1. “.”:黑色的瓷砖;

    2. “#”:白色的瓷砖;

    3. “@”:黑色的瓷砖,并且一个人站在这块瓷砖上。

    计算总共能够到达多少块黑色的瓷砖。

    #include<stdio.h>
    int x,y,i,j;
    char a[30][30];
    int f(int i,int j)
    {
    	if(i<0||j<0||i>=x||j>=y)
    	return 0;
    	if(a[i][j]=='#')
    	return 0;
    	else{
    		a[i][j]='#';
    		return 1+f(i+1,j)+f(i-1,j)+f(i,j+1)+f(i,j-1);
    	}
    }
    int main()
    {	
    	while(scanf("%d%d",&y,&x),x!=0||y!=0)
    	{
    		for(i=0;i<x;i++)
    		    scanf("%s",a[i]);
    		for(i=0;i<x;i++)
    		    for(j=0;j<y;j++)
    		        if(a[i][j]=='@')
    		            printf("%d
    ",f(i,j));
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/zyq1758043090/p/11852807.html
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