杭电ACM 1241解题报告

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转载最新推荐文章于 2025-08-09 15:58:26 发布 · 101 阅读

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原文链接：http://www.cnblogs.com/Reack/archive/2012/11/05/2755736.html

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#java

本文介绍了一个用于检测地下油藏的算法。该算法通过分析网格中每个地块来确定油藏的位置及规模，相邻的油地块被视为同一油藏的一部分。输入包括一系列网格布局，输出则是各网格中独立油藏的数量。

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Oil Deposits

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5937 Accepted Submission(s): 3461

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

代码：

#include<stdio.h>
int p[8][2]={{0,1},{0,-1},{1,0},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1}};
char ch[100][100];
int m,n;
void search(int x,int y)
{
     int i,xx,yy;
      for(i=0;i<8;i++)
     {
         xx=x+p[i][0];
         yy=y+p[i][1];
         if(xx<0||xx>=m||yy<0||yy>=n)
              continue;
         if(ch[xx][yy]=='*')
              continue;
         ch[xx][yy]='*';
         search(xx,yy);
    }
}
int main()
{
int i,j,count;
while(scanf("%d%d",&m,&n)&&m+n)
{
   for(i=0;i<m;i++)
  scanf("%s",ch[i]);
   count=0;
   for(i=0;i<m;i++)
    for(j=0;j<n;j++)
    {
     if(ch[i][j]=='@')
     {
      search(i,j);
      count++;
     }
    }
    printf("%d\n",count);
}
return 0;
}

转载于:https://www.cnblogs.com/Reack/archive/2012/11/05/2755736.html