[kuangbin带你飞]专题一简单搜索 L - Oil Deposits

最新推荐文章于 2020-07-26 09:05:10 发布

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本文介绍了一个油田探测问题，通过分析网格中的地块来确定不同油田的数量。使用深度优先搜索(DFS)算法遍历网格，标记已访问的油田地块，并计算独立油田总数。

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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.

InputThe 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.
OutputFor 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

dfs()的入门题，应该是专题一，中最水的一题了。

题目大意：给你一张地图，@代表油田*代表没有油田，当@四周（八个方向）有@则属于一片的油田，要我们计算出地图中有几片的油田。

解题思路：反正题目给的地图也很小，100*100，直接暴力寻找@，找到就进行深搜，每走一步@就把它改为*，在开个ans来记录深搜了，几次就是有几片油田。

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<string>
#include<string.h>
#include<set>
#include<math.h>
using namespace std;
int n,m;
char map[105][105];
int d[8][2]={-1,0,1,0,0,1,0,-1,1,1,-1,-1,1,-1,-1,1};
int flag;
int chark(int a,int b)///判断有没有越界
{
    return (a>=0 && a<n && b>=0 && b<m);
}
int dfs(int x,int y)
{
    if(map[x][y] == '*')
    {
        return false;
    }
    flag = 1;
    map[x][y] = '*';///每但找到一个@就改为*；
    for(int i=0; i<8; i++)
    {
        if(chark(x+d[i][0],y+d[i][1]))
        {
            dfs(x+d[i][0],y+d[i][1]);
        }
    }
    return flag;
}
int main()
{
    while(~scanf("%d%d",&n,&m) && m+n)
    {
        for(int i=0; i<n; i++)
        {
            scanf("%s",map[i]);
        }
        int ans =0;
        for(int i=0; i<n; i++) ///开始找@为入口
        {
            for(int j=0; j<m; j++)
            {
                flag = 0;
                if(dfs(i,j))
                {
                    ans ++;
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}