Lake Counting
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 41340 | Accepted: 20504 |
Description
Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.
Given a diagram of Farmer John's field, determine how many ponds he has.
Given a diagram of Farmer John's field, determine how many ponds he has.
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.
* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.
Output
* Line 1: The number of ponds in Farmer John's field.
Sample Input
10 12 W........WW. .WWW.....WWW ....WW...WW. .........WW. .........W.. ..W......W.. .W.W.....WW. W.W.W.....W. .W.W......W. ..W.......W.
Sample Output
3
Hint
OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
Source
//深度优先的典型题,大致意思是说找水坑
1 #include<iostream> 2 using namespace std; 3 char field[400][411]; 4 int n, m; 5 void dfs(int i, int j) 6 { 7 field[i][j] = '.'; 8 for(int dx=-1;dx<=1;dx++) 9 for (int dy = -1; dy <= 1; dy++) 10 { 11 int k = i + dx, kk = j + dy; 12 if (k >= 0 && k < n&& kk < m&&kk >= 0 && field[k][kk] == 'W') 13 dfs(k, kk); 14 } 15 return; 16 } 17 int main() 18 { 19 int mark = 0; 20 cin >> n >> m; 21 for (int i = 0; i < n; i++) 22 for (int j = 0; j < m; j++) 23 cin >> field[i][j]; 24 for(int i=0;i<n;i++) 25 for (int j = 0; j < m; j++) 26 if (field[i][j] == 'W') 27 { 28 dfs(i, j); 29 mark++; 30 } 31 cout << mark << endl; 32 return 0; 33 }
真正意义上的第一个深度题,代码不是很难理解,可是感觉掌握起来有点小麻烦;
还需多练习