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ACM找油田问题

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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





<h3>Sample Output</h3>





0
1
2
2

这是一个深度优先搜索的问题,先找到@然后查找他的八个方向的@,找到过的做好标记,如此递归,递归结束后结果加一。 写的代码如下:

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/************************************************************************/
/*                     SearchProbleam B                                  */
/*                     author:PSJay                                      */
/*                      date:2010-07-17                                  */
/************************************************************************/



#include "stdio.h"


int result = 0;
int land[102][102] = {0};        //声明并初始化
int flag[102][102] = {0};
int r,c;

void searchNode(int,int);

main() {
  int row,col;
  while (scanf("%d%d",&row;,&col;) != EOF) {
      
      if(row == 0 && col == 0)
          continue;
      
      //输入数据
      for (int r = 1;r <= row;r++) {
          char tempStr[101] = {'\0'};
          scanf("%s",tempStr);
          
          //转换成0,1矩阵
          for (int c = 1;c <= col;c++) {
              if(tempStr[c - 1] == '*')
                  land[r][c] = 0;
              else
                  land[r][c] = 1;
          }
      }
      
      //搜索
      for (r = 1;r <= row;r++) {
          for (c = 1;c <= col;c++) {
              if(land[r][c] == 1 && flag[r][c] == 0) {
                  searchNode(r,c);
                  result++;
              }
          }
      }
      printf("%d\n",result);
      result = 0;
      
      //还原矩阵
      for(int i = 1;i <= row;i++) {
          for(int j = 1;j <= col;j++) {
              land[i][j] = 0;
              flag[i][j] = 0;
          }
          
      }
      
  }
}

//搜索函数
void searchNode(int r,int c) {
  
  flag[r][c] = 1; //标记当前被访问的元素
  
  if(land[r-1][c-1] == 1 && flag[r-1][c-1] == 0) {
      searchNode(r-1,c-1);
  }
  if(land[r-1][c] == 1 && flag[r-1][c] == 0) {
      searchNode(r-1,c); 
  }
  if(land[r-1][c+1] == 1 && flag[r-1][c+1] == 0) {
      searchNode(r-1,c+1);
  }
  if(land[r][c+1] == 1 && flag[r][c+1] == 0) {
      searchNode(r,c+1);
  }
  if(land[r+1][c+1] == 1 && flag[r+1][c+1] == 0) {
      searchNode(r+1,c+1);    
  }
  if(land[r+1][c] == 1 && flag[r+1][c] == 0) {
      searchNode(r+1,c);
  }
  if(land[r+1][c-1] == 1 && flag[r+1][c-1] == 0) {
      searchNode(r+1,c-1);
  }
  if(land[r][c-1] == 1 && flag[r][c-1] == 0) {
      searchNode(r,c-1);
  }
}

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