小学生玩ACM----广搜

Oil Deposits

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7955    Accepted Submission(s): 4678

Problem 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

 

#include <iostream>
#include <queue>
using namespace std;
int n,m,visit[111][111],xx[8][2]={{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1}};  //visit是用来标记已经访问过了的点，xx数组是用来记录方向的，到时候就只要直接加就好了
char map[111][111];
struct ssss
{
    int x,y;  //因为要存的是坐标，没有可以装两个变量的队列，所以先放结构体打包再加入队列
}ss;
queue<ssss> q,qq;  //结构体类型的队列，q是要用到，qq是用来初始化q的
void bfs()
{
    while(!q.empty())
    {
        ss=q.front();q.pop();  //用ss取出队首元素并删除它
        int X=ss.x,Y=ss.y;
        map[X][Y]='*';visit[X][Y]=0;  //在map里面把当前坐标从@变成×，visit标记为已经访问
        for(int i=0;i<8;i++)
        {
            int x=X+xx[i][0];  //8个方向一个一个来，嘿嘿
            int y=Y+xx[i][1];
            if(x>=0&&x<n&&y>=0&&y<m)  //判断是否在范围内
                if(visit[x][y]&&map[x][y]=='@')  //判断是否访问过和是否为油田@
                {
                    visit[x][y]=0;  //标记已经访问
                    ss.x=x,ss.y=y;q.push(ss);  //先打包，再入队
                }
        }
    }
}
int main (void)
{
    int i,j,k,l,s;
    while(cin>>n>>m&&(n||m))
    {
        for(i=0;i<n;i++)
            for(j=0;j<m;j++)
                cin>>map[i][j],visit[i][j]=1;
            q=qq;
            for(i=s=0;i<n;i++)  //全图搜索
                for(j=0;j<m;j++)
                    if(map[i][j]=='@')  //当找到油田@
                        s++,ss.x=i,ss.y=j,q.push(ss),bfs();  //油田群数量加一，打包，入队，同时进行广搜，就是把它周围连着的油田全部找出来，因为这一群只能算一个油田嘛
                    cout<<s<<endl;
    }
    return 0;
}