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HDU 1312:Red and Black

2024-07-11 02:30:27   230人阅读

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8909    Accepted Submission(s): 5533


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

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>

using namespace std;

char color[25][25];
int ans;
int w, h;
int x, y;
int dx[]= {1, -1, 0, 0};
int dy[]= {0, 0, 1, -1};

void dfs(int x, int y, int w, int h)
{
    ans++;
    color[x][y]='#';
    for(int i=0; i<4; i++)
    {
        x=x+dx[i];
        y=y+dy[i];
        if(x>=0 && x<h && y>=0 && y<w && color[x][y]=='.')
            dfs(x, y, w, h);
        x-=dx[i];    //要特别注意还原数据
        y-=dy[i];
    }
}

int main()
{
    while(scanf("%d%d", &w, &h)!=EOF)
    {
        if(w==0 && h==0)
            break;
        ans=0;
        memset(color, 0, sizeof(color));
        for(int i=0; i<h; i++)
        {
            scanf("%s", color[i]);
            for(int j=0; j<w; j++)
            {
                if(color[i][j]=='@')
                {
                    x=i;
                    y=j;
                    break;
                }
            }
        }
        dfs(x, y, w, h);
        cout<<ans<<endl;
    }

    return 0;
}



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