POJ 2386 Lake Counting
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.
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.
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.
Source
USACO 2004 November
題目大意：
上下左右任意方向相鄰的W爲同一個Lake，問給定圖有多少個Lake
題目思路：
每次BFS或DFS可訪問一個W及相鄰的所有W（一個Lake），並將它們標記爲已訪問。看所有W被訪問完需要多少次搜索，即有多少個Lake
以BFS舉例：從任意的一個W開始BFS，搜索到一個W就把它設置爲非W，直至此次BFS結束，Lake的數量加1；再從剩餘的W中進行BFS，同樣操作至結束，Lake數量加1；當圖中不存在W時，Lake即爲所求。
同樣原理的題目，這篇博客HDU1241，解法就是BFS。
但是此處採用DFS也可以解決
代碼：
#include<iostream>
#include<cstdlib>
#include<string>
using namespace std;
int n, m;
char lake[101][101];
void dfs(int i,int j)
{
lake[i][j] = '.';
for (int p = -1; p<= 1; p++)
{
for (int q = -1; q <= 1; q++)
{
int a = i + p, b = j + q;
if (a >= 1 && a <= n && b >= 1 && b <= m && lake[a][b] == 'W') dfs(a, b);
}
}
}
int main()
{
int sum = 0;
cin >> n >> m;
for (int i = 1; i <= n; i++)
{
string s;
cin >> s;
for (int j = 0; j < m; j++)
{
if (s[j] == '.')lake[i][j + 1] = '.';
else lake[i][j + 1] = 'W';
}
}
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
if (lake[i][j] == 'W') {
dfs(i,j);
sum++;
}
cout << sum << endl;
return 0;
}
