HDU-1241-Oil Deposits（BFS）

J - Oil Deposits

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit

Status

Practice

HDU 1241

Appoint description:

Description

The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil. A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid.

Input

The input file contains one or more grids. Each grid begins with a line containing m and n, the number of rows and columns in the grid, separated by a single space. If m = 0 it signals the end of the input; otherwise 1 <= m <= 100 and 1 <= n <= 100. Following this are m lines of n characters each (not counting the end-of-line characters). Each character corresponds to one plot, and is either

*', representing the absence of oil, or

@’, representing an oil pocket.

Output

For each grid, output the number of distinct oil deposits. Two different pockets are part of the same oil deposit if they are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets.

Sample Input

1 1

*

3 5

@@*

@

@@*

1 8

@@**@*

5 5

**@

@@@

@*@

@@@*@

@@**@

0 0

Sample Output

0

1

2

2

题意：@代表油井，横纵斜八个方向都算是相邻，问有多少片油井是独立存在的。

思路：遍历图，遇到@就把它八个方向的油井都标记，然后入队循环，直至队列为空，这时独立数量++即可

代码

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<string.h>
using namespace std;
const int maxn=105;
char map[maxn][maxn];//存图
int dis[8][2]= {{0,1},{0,-1},{1,0},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1}}; //八个搜素方向
int M;//M行
int N;//N列
int num;//已找到油田片的数量
struct node
{
int x;
int y;
};
bool Judge(int x,int y)
{
if(x>=0&&x<M&&y>=0&&y<N&&map[x][y]=='@')
return true;
return false;
}
void BFS()
{
num=0;//初始化油田连片数量
for(int i=0; i<M; i++)
{
for(int j=0; j<N; j++)
{
if(map[i][j]=='@')//遍历图找到油井
{
queue<node>q;
node star;
star.x=i;
star.y=j;
q.push(star);
while(!q.empty())
{
star=q.front();
q.pop();
for(int i=0; i<8; i++) //八个搜索方向
{
node end=star;
end.x+=dis[i][0];
end.y+=dis[i][1];
if(Judge(end.x,end.y))//和它相连的油井全部设置为*
{
map[end.x][end.y]='*';
q.push(end);
}
}
}
num++;
}
}
}
printf("%d\n",num);
}
void init()
{
for(int i=0; i<M; i++)
for(int j=0; j<N; j++)
cin>>map[i][j];
}
int main()
{
while(~scanf("%d%d",&M,&N))
{
if(N==0&&M==0)
break;
init();
BFS();
}
return 0;
}