uva 572 oil Deposits-优快云博客

本文链接：https://blog.youkuaiyun.com/u012315428/article/details/17218095

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的数据来确定油藏的位置，并能识别出不同的油藏沉积物。输入包括网格大小及地块是否含有油的信息，输出则是独立油藏的数量。

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Problem A : Oil Deposits

Oil Deposits

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 $1 \le m \le 100$ and $1 \le n \le 100$ . 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

#include<iostream>
#include<vector>
#include<string>
using namespace std;
/////////////////
const int dx[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
const int dy[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
class OilDeposits{
private:
vector<string> land;
int numberOfDeposits;
void dfs(int x, int y);
public:
void initial(){land.clear(); numberOfDeposits = 0;}
void readCase(int m, int n);
void computing();
void outResult(){cout << numberOfDeposits << endl;}
void outLand(){
for(int i = 0; i < land.size(); i++){
cout << land[i] << endl;
}
cout << endl;
}
};
void OilDeposits::readCase(int m, int n){
land.resize(m + 2);
for(int i = 0; i < n + 2; i++){
land[0].push_back('*'); //add the top boundary
land[m+1].push_back('*'); //add the bottom boundary
}
for(int i = 1; i <= m; i++){
cin >> land[i];
land[i] = '*' + land[i] + '*'; //add the left and right boundaries
}
}
void OilDeposits::computing(){
for(int r = 1; r < land.size() - 1; r++){
for(int c = 1; c < land[r].size() - 1; c++){
if(land[r][c] == '@'){
numberOfDeposits++;
dfs(r, c);
}
}
}
}
void OilDeposits::dfs(int x, int y){
if(land[x][y] == '@'){
land[x][y] = '*';
for(int i = 0; i < 8; i++){
dfs(x + dx[i], y + dy[i]);
}
}
}
///////////////
int main(){
OilDeposits od;
int m, n;
while((cin >> m >> n) && m){
od.initial();
od.readCase(m, n);
//od.outLand();
od.computing();
od.outResult();
}
return 0;
}
/*
INPUT
1 1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5
****@
*@@*@
*@**@
@@@*@
@@**@
0 0
------------------
OUTPUT
0
1
2
2
*/