Red and Black

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. 

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

Source

Asia 2004, Ehime (Japan), Japan Domestic

原本代码：2014.5.24

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int D,SIGN,Len_X,Len_Y;
char Map[100][100];
int Sign[100][100];

int Sign_Part(int x,int y)
{

    int ii,jj;
    if(x>=Len_X||x<0||y>=Len_Y||y<0)return 0;
    if(Map[x][y]=='#')return 0;
    else Map[x][y]='#';
    /*putchar('
');
       for(ii=0;ii<Len_X;ii++)
        {
            for(jj=0;jj<Len_Y;jj++)
                printf("%c",Map[ii][jj]);
            putchar('
');
        }*/
    Sign[x][y]=1;
    SIGN++;
    if(Map[x+1][y]!='#'&&x+1<Len_X&&Sign[x+1][y]!=1)
    {
        Sign_Part(x+1,y);
        Map[x+1][y]='.';
    }
    if(Map[x-1][y]!='#'&&x-1>=0&&Sign[x-1][y]!=1)
    {
        Sign_Part(x-1,y);
        Map[x-1][y]='.';
    }
    if(Map[x][y+1]!='#'&&y+1<Len_Y&&Sign[x][y+1]!=1)
    {
        Sign_Part(x,y+1);
        Map[x][y+1]='.';
    }
    if(Map[x][y-1]!='#'&&y-1>=0&&Sign[x][y-1]!=1)
    {
        Sign_Part(x,y-1);
        Map[x][y-1]='.';
    }
    return 0;

}

int main()
{
    int i,j,Begin_x,Begin_y;
    while(scanf("%d%d",&Len_Y,&Len_X)!=EOF)
    {
        if(Len_X==0&&Len_Y==0)break;
        for(i=0;i<Len_X;i++)
        {
            getchar();
            for(j=0;j<Len_Y;j++)
            {
                scanf("%c",&Map[i][j]);
                if(Map[i][j]=='@')
                {Begin_x=i;Begin_y=j;}
            }
        }
        SIGN=0;
        memset(Sign,0,sizeof(Sign));
        Sign_Part(Begin_x,Begin_y);
        printf("%d
",SIGN);

    }

    return 0;
}

修改：2015.5.8

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
using namespace std;
int D,SIGN,Len_X,Len_Y;
char Map[100][100];
void BFS(int x,int y)
{
    if(Map[x][y]=='#')return ;
    else
    {
        SIGN++;
        Map[x][y]='#';
        BFS(x-1,y);
        BFS(x+1,y);
        BFS(x,y-1);
        BFS(x,y+1);
    }
}

int main()
{
    int i,j,Begin_x,Begin_y;
    while(scanf("%d%d",&Len_Y,&Len_X)!=EOF)
    {
        if(Len_X==0&&Len_Y==0)break;
        for(i=0;i<=Len_X+1;i++)
        {
            for(j=0;j<=Len_Y+1;j++)
            {
                if(i==0||j==0||i==Len_X+1||j==Len_Y+1)Map[i][j]='#';
                else
                {
                    scanf(" %c",&Map[i][j]);
                    if(Map[i][j]=='@')
                    {Begin_x=i;Begin_y=j;}
                }

            }
        }
        SIGN=0;
        BFS(Begin_x,Begin_y);
        printf("%d
",SIGN);

    }
    return 0;
}