1241 Oil Deposits

最新推荐文章于 2021-01-24 13:36:02 发布

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本文介绍了一个算法挑战，任务是分析网格地图中石油沉积的数量。通过深度优先搜索（DFS）算法，遍历每个网格单元确定独立的石油沉积区，并计算总的沉积数量。此问题涉及网格遍历、递归调用及标记已访问区域的技术。

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Oil Deposits

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15592 Accepted Submission(s): 8943

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

水题。。。。但是没有看清楚题目的意思，刚开始以为是4个方向，结果是8个方向，然后脑残的用set来存，会有出现两个不一样的油田但是油数量一样的情况，因此WA了，最后还是老老实实用填充的方法，遇到一个'@'就遍历填充，sum++，直到全部遍历

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <vector>
#include <algorithm>
#include <stdlib.h>
#include <set>
using namespace std;
#define M 105
char map[M][M];
int n, m;
set <int> se;
int dir[][2] = {1,0,-1,0,0,1,0,-1,1,1,1,-1,-1,-1,-1,1};
int sum;
void DFS(int x, int y){
 int num1, num2;
 for(int i = 0; i < 8; i++){
  num1 = x + dir[i][0];
  num2 = y + dir[i][1];
  if(map[num1][num2] == '@'){
   map[num1][num2] = '*';
   DFS(num1,num2);
  }
 }
}
int main(){
 while(scanf("%d%d",&n,&m)!=EOF){
  if(n == 0 && m == 0)
   break;
  memset(map,0,sizeof(map));
  for(int i = 0; i < n; i++){
   scanf("%s",&map[i]);
  }
  sum = 0;
  for(int i = 0; i < n; i++){
   for(int j = 0; j < m; j++){
    if(map[i][j] == '@'){
     map[i][j] = '*';
     DFS(i,j);
     sum ++;
    }
   }
  }
  printf("%d\n",sum);
  se.clear();
 }
    return 0;
}