题目来源:http://poj.org/problem?id=1562
之前在hdoj上写了一个基于vector的深搜点击打开链接,当时是为了和宽搜对比,都用了vector而没有用递归。其实递归的深搜思路更清晰更具有可扩展性,所以在这里又写了一个递归的深搜。
Oil Deposits
Time Limit: 1000MS |
Memory Limit: 10000K |
Description
The GeoSurvComp geologic survey company isresponsible for detecting underground oil deposits. GeoSurvComp works with onelarge rectangular region of land at a time, and creates a grid that divides theland into numerous square plots. It then analyzes each plot separately, usingsensing equipment to determine whether or not the plot contains oil. A plotcontaining oil is called a pocket. If two pockets are adjacent, then they arepart of the same oil deposit. Oil deposits can be quite large and may contain numerouspockets. Your job is to determine how many different oil deposits are containedin a grid.
Input
Theinput contains one or more grids. Each grid begins with a line containing m andn, the number of rows and columns in the grid, separated by a single space. Ifm = 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 (notcounting the end-of-line characters). Each character corresponds to one plot,and is either `*', representing the absence of oil, or `@', representing an oilpocket.
Output
are adjacent horizontally, vertically, ordiagonally. 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
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解题思路
递归深搜
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代码
//Oil Deposits //Time Limit: 1000MS Memory Limit: 10000K //Total Submissions: 20679 Accepted: 10808 //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 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 // //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 #include<fstream> #include<iostream> using namespace std; const int MAX = 101; char mat[MAX][MAX]; void search(int i, int j, int m, int n) { mat[i][j] = '$'; if (i>0 && mat[i-1][j]=='@') { search(i-1,j,m,n); } if (j>0 && mat[i][j-1]=='@') { search(i,j-1,m,n); } if (i>0 && j>0 && mat[i-1][j-1]=='@') { search(i-1,j-1,m,n); } if (i<m-1 && mat[i+1][j]=='@') { search(i+1,j,m,n); } if (j<n-1 && mat[i][j+1]=='@') { search(i,j+1,m,n); } if (i<m-1 && j<n-1 && mat[i+1][j+1]=='@') { search(i+1,j+1,m,n); } if (i>0 && j<n-1 && mat[i-1][j+1]=='@') { search(i-1,j+1,m,n); } if (i<m-1 && j>0 && mat[i+1][j-1]=='@') { search(i+1,j-1,m,n); } } int main() { #ifndef ONLINE_JUDGE ifstream fin("poj1562.txt"); int m,n,i,j,cnt=0; while(fin >> m >> n) { if (m==0) { break; } cnt = 0; for (i=0; i<m; i++) { for (j=0; j<n; j++) { fin >> mat[i][j]; } } for (i=0; i<m; i++) { for (j=0; j<n; j++) { if (mat[i][j] == '@') { search(i,j,m,n); cnt ++; } } } cout << cnt << endl; } #endif #ifdef ONLINE_JUDGE int m,n,i,j,cnt=0; while(cin >> m >> n) { if (m==0) { break; } cnt = 0; for (i=0; i<m; i++) { for (j=0; j<n; j++) { cin >> mat[i][j]; } } for (i=0; i<m; i++) { for (j=0; j<n; j++) { if (mat[i][j] == '@') { search(i,j,m,n); cnt ++; } } } cout << cnt << endl; } #endif // ONLINE_JUDGE }