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Lake Counting -- DFS（深搜）

2015-04-19 11:33 169 查看

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 22424		Accepted: 11300

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

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

解题分析：

首先这道题目设计几个问题：

第一个就是怎么可以确定几个w是属于一块的。

第二个怎么搜索。

第三个怎么确保不会搜索重复的。

那么第一个和第二个是一个问题，就是深度优先搜索，然后，利用递归的方法，每次对于一个点是否有8个方向，如果有的话，继续8个方向的搜索，直到到达边界为止。

还有怎么确定不重复，那么就是我把每次找到的w都变味' . ' 就可以了。

具体看代码。

代码：

#include <iostream>

#include <cstdio>

#define MAXN 102

using namespace std;

char map[MAXN][MAXN];

int N, M;

int DG(int i, int j) {

if(i < 0 || i >= N || j < 0 || j >= M)

return 0;

if(map[i][j] == '.')

return 0;

map[i][j] = '.';

return 1 + DG(i-1, j-1) + DG(i-1, j) + DG(i-1, j+1)

+ DG(i, j-1) + DG(i, j+1) + DG(i+1, j-1) +

DG(i+1, j) + DG(i+1, j+1);

}

int main()

{

cin >> n >> m;

char flag;

for(int i = 0; i < n; i++) {

for(int j = 0; j < m; j++) {

cin >> flag;

map[i][j] = flag;

}

}

int count = 0;

for(int i = 0; i < n; i++) {

for(int j = 0; j < m; j++) {

if(DG(i, j) != 0)

count++;

}

}

cout << count << endl;

return 0;

}

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