Red and Black（BFS）

ゝ一世哀愁。 2022-05-19 07:10 121阅读 0赞

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

思路：本题是简单的深搜bfs模板

MY DaiMa：

#include<iostream>
    #include<stdio.h>
    #include<queue>
    #include<string.h>
    using namespace std;
    int dir[4][2]={
       {0,1},{0,-1},{1,0},{-1,0}};
    int n,m,t;
    char flag[21][21];
    struct point
    {
        int x,y;
    }start;
    int bfs()
    {
        queue<point>Q;
        point cur,next;
        flag[start.x][start.y]='#';
        Q.push(start);
        while(!Q.empty())
        {
            cur=Q.front();
            Q.pop();
            for(int i=0;i<4;i++)
            {
                next.x=cur.x+dir[i][0];
                next.y=cur.y+dir[i][1];
                if((next.x>=0&&next.x<n)&&(next.y>=0&&next.y<m)&&flag[next.x][next.y]=='.')
                {
                    t++;
                    flag[next.x][next.y]='#';
                    Q.push(next);
                }
            }
        }
        return t;
    }
    int main()
    {
        while(cin>>m>>n&&(n!=0&&m!=0))
        {
            t=1;
            for(int i=0;i<n;i++)
            {
                for(int j=0;j<m;j++)
                {
                    cin >> flag[i][j];
                    if(flag[i][j]=='@')
                    {
                        start.x=i;
                        start.y=j;
                    }
                }
            }
            //t = bfs();
            cout << bfs() <<endl;
        }
        return 0;
    }