一道简单的搜索题，和poj2386差不多，DFS和BFS都可以,当然并查集也可以……
Red and Black
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 14518 | Accepted: 7537 |
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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
BFS；
#include<iostream>
#include<cstring>
using namespace std;
struct Node
{
int x;
int y;
}queue[7000];
char map[21][21];
int visited[21][21];
int w, h;
int ax[4]={-1, 1, 0, 0};
int ay[4]={0, 0, -1, 1};
int BFS(int x, int y)
{
int num=1;//初始位置为黑色，所以初始值为1
int front=0, rear=1;
int i;
int X,Y;
visited[y][x]=1;//标记
queue[front].x=x;
queue[front].y=y;
while( front != rear)
{
for(i=0; i<4; i++)
{
X=queue[front].x+ax[i];
Y=queue[front].y+ay[i];
if( !visited[Y][X] && (X>=0&&X<w) && (Y>=0&&Y<h))//未越界且未访问过，入队列
{
queue[rear].x=X;
queue[rear].y=Y;
visited[Y][X]=1;
rear++;
num++;
}
}
front++;
}
return num;
}
int main()
{
int i, j;
int x, y;
while( cin>>w>>h && w)
{
memset(visited, 0, sizeof(visited));
for(i=0; i<h; i++)
for(j=0; j<w; j++)
{
cin>>map[i][j];
if( map[i][j]=='#')
visited[i][j]=1;
else if( map[i][j]=='@')//记录人的初始位置
{
x=j;y=i;
}
}
cout<<BFS(x, y)<<endl;
}
}
DFS:
#include<iostream>
#include<cstring>
using namespace std;
char map[21][21];
int visited[21][21];
int w, h;
int num;//初始位置为黑色，所以初始值为1
int ax[4]={-1, 1, 0, 0};
int ay[4]={0, 0, -1, 1};
void DFS(int x, int y)
{
int i;
if( visited[y][x] == 1 || (x<0||x>=w) || (y<0||y>=h)) //访问过，或越界，递归结束
return ;
else//未访问过,继续
{
visited[y][x]=1;//标记
num++;
for(i=0; i<4; i++)
{
DFS(x+ax[i], y+ay[i]);
}
}
}
int main()
{
int i, j;
int x, y;
while( cin>>w>>h && w)
{
num=0;//初始化初始值
memset(visited, 0, sizeof(visited));
for(i=0; i<h; i++)
for(j=0; j<w; j++)
{
cin>>map[i][j];
if( map[i][j]=='#')
visited[i][j]=1;
else if( map[i][j]=='@')//记录人的初始位置
{
x=j;y=i;
}
}
DFS(x, y);
cout<<num<<endl;
}
}