杭电1241 Oil Deposits（搜索题）

Oil Deposits

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19973    Accepted Submission(s): 11475

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

看题目的输入方式不用读题就能知道这是一道搜索题目0.0

至于读者的你想用广搜还是深搜那就取决于您自己的风格了

我这里贴上我的广搜AC代码~

#include
#include
#include
using namespace std;
struct zuobiao
{
    int x,y;
}now,nex;
int fx[8]={-1,-1,0,1,1,1,0,-1};
int fy[8]={0,1,1,1,0,-1,-1,-1};
char a[110][110];
int vis[110][110];
    int m,n;
    int output;
void bfs(int x,int y)
{
    if(vis[x][y]==0)output++;
    vis[x][y]=1;
    queues;
    now.x=x;
    now.y=y;
    s.push(now);
    while(!s.empty())
    {
        now=s.front();
        s.pop();
        for(int i=0;i<8;i++)
        {
            nex.x=now.x+fx[i];
            nex.y=now.y+fy[i];
            if(nex.x>=0&&nex.x=0&&nex.y             {
                vis[nex.x][nex.y]=1;
                s.push(nex);
            }
        }
    }
}
int main()
{
    while(~scanf("%d%d",&m,&n))
    {
        if(m==0&&n==0)break;
        output=0;
        memset(vis,0,sizeof(vis));
        memset(a,0,sizeof(a));
        for(int i=0;i         {
            scanf("%s",a[i]);
        }
        for(int i=0;i         {
            for(int j=0;j             {
                if(a[i][j]=='@'&&vis[i][j]==0)
                bfs(i,j);
            }
        }
        printf("%d\n",output);
    }
}