[kuangbin带你飞]专题一简单搜索L - Oil Deposits(HDU 1241)

最新推荐文章于 2020-09-28 01:36:15 发布

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本文介绍了一种用于探测地下油田分布情况的算法。通过建立矩形网格并使用传感设备确定每个地块是否含有石油，进而识别不同油田的数量。算法采用广度优先搜索（BFS），确保相邻的石油地块被视为同一油田。

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L - Oil Deposits

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1241

Appoint description:

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

题意：求有几个油田，八个方向有一个方向是相同的就是联通的，也就是一个。

直接bfs

跟图的遍历差不多

遍历后把@改为*（判重）

#include
  
   
#include
   
    
#include
    
     
#include
     
      
using namespace std;

const int N = 1000;
int n, m;
char str[N][N];

struct node
{
    int x, y;
    node(){}
    node(int xx, int yy)
    {
        x = xx; y = yy;
    }
};

queue
      
       que;

int dir[9][2] = { {1, 0}, {-1, 0}, {0, 1}, {0, -1}, {1, 1}, {-1, -1}, {1, -1}, {-1, 1} };

int bfs(int x, int y)
{
    while(!que.empty()) que.pop();
    que.push( node(x, y) );
    while(!que.empty())
    {
        node tmp = que.front(); que.pop();
        int xx, yy;
        for(int i = 0; i < 8; i++)
        {
            xx = tmp.x + dir[i][0];
            yy = tmp.y + dir[i][1];
            if(0 <= x && x < n && 0 <= yy && yy < m && str[xx][yy] == '@')
            {
                str[xx][yy] = '*';
                que.push( node(xx, yy) );
            }
        }
    }
    return 0;
}

int main(void)
{
    while(scanf("%d%d", &n, &m), n+m)
    {
        int i, j;
        for(i = 0; i < n; i++)
            scanf("%s", str[i]);
        int ans = 0;
        for(i = 0; i < n; i++)
            for(j = 0; j < m; j++)
            {
                if(str[i][j] == '@')
                {
                    bfs(i, j);
                    ans++;
                }
            }
        printf("%d\n", ans);
    }
    return 0;
}