POJ 1979Red and Black

Posted 2020-09-09 九月旧约

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 34862		Accepted: 18856

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) 
The end of the input is indicated by a line consisting of two zeros. 

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
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#[email protected]#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
[email protected]
###.###
..#.#..
..#.#..
0 0

Sample Output

45
59
6
13

Source

Japan 2004 Domestic

#include <stdio.h>
#include <math.h>
#include <iostream>
#include <string.h>
#include <queue>
#define maxn 105
using namespace std;
int n,m,num,vis[maxn][maxn];
int dx[]={1,-1,0,0},dy[]={0,0,1,-1};
char mapn[maxn][maxn];
struct node
{
    int x,y;
};
void bfs(int a,int b)
{
    queue<node> q;
    node p;
    p.x = a;
    p.y = b;
    num++;
    vis[a][b] = 1;
    q.push(p);
    while(!q.empty())//加入队列的都是可以走到的点，所以只需ans++即可
    {
        node now = q.front();
        int x = now.x,y = now.y;
        q.pop();
        for(int i=0;i<4;i++)
        {
            int xx = x + dx[i];
            int yy = y + dy[i];
            if(xx<0||yy<0||xx>(n-1)||yy>(m-1)||mapn[xx][yy]==‘#‘||vis[xx][yy])
                 continue;
            num++;
            node tmp;
            tmp.x = xx;
            tmp.y = yy;
            q.push(tmp);
            vis[xx][yy] = 1;
        }
    }
}
int main()
{
    while(cin >> m >> n)//注意这里输入的是m,n
    {
        memset(vis,0,sizeof(vis));
        num = 0;
        if(n==0||m==0)
            break;
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
                cin >> mapn[i][j];
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
        {
            if(mapn[i][j]==‘@‘)
                bfs(i,j);
        }
        cout << num << endl;
    }
}