搜索 P题

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

 0122 

2.解题思路：

    题意：给定一幅地图寻找油矿，去判断相邻，对角的油田有几个。@代表油田，这个点的上下左右斜角上是“@”的点属于同一个油矿。

  思路：本题为DFS题，当搜到一个@的时候对上下左右及四个对角进行深搜，将相邻的@都进行标记，表示已经搜索过，把所有数据搜索完就能统计出油田的个数了。
3.代码：

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
char map[101][101];
int n, m, p;
void dfs(int i, int j)
{
    if(map[i][j]!='@' || i<0 || j<0 || i>=m || j>=n) return;
    else
    {
        map[i][j]='*';
        dfs(i-1, j-1);
        dfs(i-1, j);
        dfs(i-1, j+1);
        dfs(i, j-1);
        dfs(i, j+1);
        dfs(i+1, j-1);
        dfs(i+1, j);
        dfs(i+1, j+1);
    }
}
int main()
{
    int i, j;
    while(cin>>m>>n&&m!=0)
    {
        if(m==0||n==0) break;
        p = 0;
        for(i = 0; i < m; i++)
        for(j = 0; j < n; j++)
        cin>>map[i][j];
        for(i = 0; i < m; i++)
        {
        for(j = 0; j < n; j++)
        {
        if(map[i][j] == '@')
        {
            dfs(i, j);
            p++;
        }
        }
        }
        cout<<p<<endl;
    }


    return 0;
}