【HDU-1241】Oil Deposits油田（dfs）_hdu dfs 油田-优快云博客

本文链接：https://blog.youkuaiyun.com/iceiceicpc/article/details/50830052

本文介绍了一个基于简单深度优先搜索(DFS)算法来解决油藏探测问题的方法。该算法用于确定给定网格中独立油藏的数量，通过遍历每个网格单元并标记已访问过的油藏区域实现。适用于地质勘探公司对地下油藏进行快速有效的评估。

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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

题目大意

给你一个图，求互不相交的联通块的个数

思路

简单的dfs，没搜索一个联通块总数count+1

代码

#include <iostream>
#include <math.h>
using namespace std;
int dire[2][8]={{1,-1,0,0,1,-1,1,-1},{0,0,1,-1,1,-1,-1,1}};
int n,m;
char oil[110][110];
int chess[110][110];
int OilCount;
void bfs(int x,int y)
{
    int i,desX,desY;
    for(int i=0;i<8;i++)
    {
        desX=x+dire[0][i];
        desY=y+dire[1][i];
        if(oil[desX][desY]=='@'&&desX>=1&&desX<=n&&desY>=1&&desY<=m&&!chess[desX][desY])
        {
            /*if(oil[desX][desY]=='@') 
            {
            checkX[settime]=desX;
            checkY[settime]=desY;
            if(!IsLeagal(desX,desY)) OilCount++;
            settime++;
            }*/
            chess[desX][desY]=1;
            bfs(desX,desY);
        }
    }
}


int main()
{
    while(scanf("%d %d",&n,&m)!=EOF&&(n||m))
    {

        memset(chess,0,sizeof(chess));
        /*for(int i=1;i<=n;i++)
        {
            getchar();
            for(int j=1;j<=m;j++)
            {
                scanf("%c",&oil[i][j]);
            }
        }*/
        for( int i = 1; i <= n ;i++ )
            scanf("%s",oil[i]+1);
        //gets(oil[i]+1);
        OilCount=0;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
            {
                if(oil[i][j]=='@'&&!chess[i][j])
                {
                    OilCount++;
                    bfs(i,j);
                }
            }
        printf("%d\n",OilCount);
    }
    return 0;
}