HDU 1241 Oil Deposits（DFS）

油田探测计数

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本文介绍了一个经典的图遍历问题——油田探测计数。通过深度优先搜索(DFS)的方法，实现对给定区域中油田数量的有效计数。每发现一个新的油田就进行一次深度优先遍历，并标记已访问过的油田，避免重复计数。

Oil Deposits

原题链接：HDU 1241

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

在这里插入图片描述

Sample Output

0
1
2
2

题解：

题意：

有一块地，判断油田的个数，@代表油田，如果几块油田相邻则表示一块油田。

AC代码

#include<iostream>
#include<string.h>

using namespace std;
char a[105][105];
int ne[105][105];
int m,n;


void dfs(int x,int y,int num) {
    if(x >= m || y >= n || x < 0 || y  < 0)
        return;
    if(ne[x][y] > 0 || a[x][y] != '@')
        return;
    ne[x][y] = num;
    for(int i = -1; i <= 1; i++)
        for(int j = -1; j <= 1; j++)
            dfs(x+i,y+j,num);
}
int main() {
    std::ios::sync_with_stdio(false);

    while(cin>>m>>n && m != 0) {
        memset(ne,0,sizeof(ne));
        int cunt = 0;
        for(int i = 0; i < m; i++)
            cin>>a[i];
        for(int i = 0; i < m; i++)
            for(int j = 0; j  < n; j++)
                if(ne[i][j] == 0 && a[i][j]=='@')
                    dfs(i,j,++cunt);
        cout<<cunt<<endl;
    }
    return 0;
}