1312 Red and Black

#問題

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25498    Accepted Submission(s): 15379

 

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

 

 

#分析

這是一道圖的遍歷題，我選擇使用DFS，記錄所有到達點的數量即可

#AC碼

#include<iostream>
#include<stdio.h>
#include<string.h>
//point:0可走 -1不可走
//c_cn:0未計數 1已計數
//c_dfs:dfs過程中標記使用
using namespace std;
int cn;
class _Map
{
public:
    int point;
    int c_dfs;
    int c_cn;
    _Map()
    {
        point=0;
        c_cn=0;
        c_dfs=0;
    }
}Map[22][22];
void DFS(int x,int y)
{
    Map[x][y].c_dfs=1;
    if(!Map[x][y].c_cn)
    {
        cn++;
        Map[x][y].c_cn=1;

    }
    else       //這裏return很有必要，即遇見大量.的時候可以不重複計量
        return;
    if(!Map[x+1][y].point&&!Map[x+1][y].c_dfs)
    {
        DFS(x+1,y);
        Map[x+1][y].c_dfs=0;
    }
    if(!Map[x-1][y].point&&!Map[x-1][y].c_dfs)
    {
        DFS(x-1,y);
        Map[x-1][y].c_dfs=0;
    }
    if(!Map[x][y+1].point&&!Map[x][y+1].c_dfs)
    {
        DFS(x,y+1);
        Map[x][y+1].c_dfs=0;
    }
    if(!Map[x][y-1].point&&!Map[x][y-1].c_dfs)
    {
        DFS(x,y-1);
        Map[x][y-1].c_dfs=0;
    }
    return;
}
int main()
{
    int x,y;
    int cx,cy;
    //freopen("1312.txt","r",stdin);
    while(cin>>y>>x,x)
    {
        cn=0;
        memset(Map,0,sizeof(Map));
        //設置邊界
        for(int i=0;i<=y;i++)
        {
            Map[0][i].point=-1;
            Map[x+1][i].point=-1;
        }
        for(int i=0;i<=x;i++)
        {
            Map[i][0].point=-1;
            Map[i][y+1].point=-1;
        }
        //錄入地圖
        for(int i=1;i<=x;i++)
        for(int j=1;j<=y;j++)
        {
            char ch;
            cin>>ch;
            if(ch=='#')
                Map[i][j].point=-1;
            if(ch=='@')
            {
                cx=i;
                cy=j;
            }
        }
        DFS(cx,cy);
        cout<<cn<<endl;
    }
}