HDU 1312 Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

BFS

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

题意：

一块矩形的地面，贴有红色和黑色的地砖，一个人只能走黑色地砖，能走所在地砖的上，下，左，右四个方向的地砖，给你一个图问总共可以走多少块地砖

解题思路：

一个裸的BFS

完整代码：

#include<bits/stdc++.h>
using namespace std;
int dir[4][2]={0,1,0,-1,1,0,-1,0};///方向;
int vis[25][25];///标记(走过为1，没走为0);
char a[25][25];///存图;
int m,n,num;///m列，n行，走过num个点;
///结构体存点的坐标:
struct node
{
    int x;
    int y;
};
///从起点开始搜:
int bfs(int x,int y)
{
    queue<node>q;
    node now,next;
    now.x=x;
    now.y=y;
    q.push(now);
    num++;
    vis[x][y]=1;
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        for(int i=0;i<4;i++)
        {
            int nx=now.x+dir[i][0];
            int ny=now.y+dir[i][1];
            if(a[nx][ny]=='.'&&vis[nx][ny]==0&&nx>=0&&nx<n&&ny>=0&&ny<m)
            {
                next.x=nx;
                next.y=ny;
                num++;
                vis[nx][ny]=1;
                q.push(next);
            }
        }
    }
    return num;
}
int main()
{
    int x,y;
    while(scanf("%d%d",&m,&n),m+n)
    {
        num=0;
        memset(a,0,sizeof(a));
        memset(vis,0,sizeof(vis));
        for(int i=0;i<n;i++)
            scanf("%s",a[i]);
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
            {
                ///找起点:
                if(a[i][j]=='@')
                {
                    x=i;
                    y=j;
                }
            }
        printf("%d\n",bfs(x,y));
    }
    return 0;
}