DFS和BFS例题

FuBin29 · Apr 15, 2020 · 119df30 · 119df30
1 parent 3a7daed
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diff --git a/Algorithm.sln b/Algorithm.sln
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diff --git a/Lake_Counting_POJ2386-DFS/Lake_Counting_POJ2386-DFS.vcxproj b/Lake_Counting_POJ2386-DFS/Lake_Counting_POJ2386-DFS.vcxproj
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diff --git a/Lake_Counting_POJ2386-DFS/Lake_Counting_POJ2386-DFS.vcxproj.filters b/Lake_Counting_POJ2386-DFS/Lake_Counting_POJ2386-DFS.vcxproj.filters
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diff --git a/Lake_Counting_POJ2386-DFS/main.cpp b/Lake_Counting_POJ2386-DFS/main.cpp
@@ -0,0 +1,46 @@
+#include<iostream>
+
+using namespace std;
+
+//深度优先搜索的方式来遍历一个水坑周围的所有地点，将水坑变为空地
+void dfs(char field[][101], int i, int j, int N, int M)
+{
+	//将水坑变为空地
+	field[i][j] = '.';
+
+	//遍历8个方向
+	for (int dx = -1;dx<=1;dx++)
+		for (int dy = -1; dy <= 1; dy++)
+		{
+			int nx = i + dx;
+			int ny = j + dy;
+			if (nx >= 1 && nx <= N && ny >= 1 && ny <= M && field[nx][ny] == 'W')
+				dfs(field, nx, ny, N, M);
+		}
+	return;
+}
+
+int main()
+{
+	int N, M;
+	char field[101][101] = { '\0' };
+	cin >> N >> M;
+	for (int i = 1; i <= N; i++)
+		for (int j = 1; j <= M; j++)
+			cin >> field[i][j];
+
+	int lake = 0;
+	//遍历所有水坑，有点水坑会在DFS过程中由于合并的需要被改写为空地
+	for (int i = 1; i <= N; i++)
+		for (int j = 1; j <= M; j++)
+			if(field[i][j] == 'W')
+			{
+				//使用DFS遍历一个水坑，及其相连的水坑
+				dfs(field, i, j, N, M);
+				lake++;
+			}
+
+	cout << lake << endl;
+
+	return 0;
+}
diff --git a/Lake_Counting_POJ2386-DFS/问题.md b/Lake_Counting_POJ2386-DFS/问题.md
@@ -0,0 +1,49 @@
+# Lake Counting POJ 2386
+
+http://poj.org/problem?id=2386
+
+### 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.
+
+关于水坑的相连：一个W周围的8个位置如果也是W，则可以连起来，算作一个水坑。因此想到了用穷竭搜索的方式来统计一个水坑，这里选择了用DFS，因为好写程序，栈的形式和递归函数调用很相似，如果用BFS，还要想存储结构。
+
+### 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.
diff --git a/Readme.md b/Readme.md
@@ -3,13 +3,16 @@
 | 项目名 | 描述 |
 | ------ | ---- |
 |平面最近点对-DC| 二维平面点集，寻找距离最近的两个点，使用分治法一分为二的解决 |
-|01背包-DP||
+|01背包-DP|01背包问题，用DP解决，有多种DP格式|
 |钢条切割-DP||
 |最优二叉搜索树-DP||
 |最长单调递增子序列-DP||
+|Lake_Counting_POJ2386-DFS|二维字符数组，一个W周围8个空间有W，则相连为一个水坑，对水坑计数。|
 |Eight_Puzzle-BFS||
+|迷宫最短路径-BFS|在一个迷宫里从起点到终点最少走多少步|
 |Nqueens-局部快速搜索||
 |木板切割开销_POJ3235-贪心Huffman|切割一块木板的花费等于木板长度，给定切割结果，求最小花费，等同于Huffman问题，用贪心算法求解|
 |士兵布置_POJ3069-贪心|在一维数轴上，每个士兵有一定控制范围，要求给定的若干点必须被士兵控制，问最小士兵数，从数轴的一端开始用贪心算法放置士兵|
 |字典序最小问题_POJ3617-贪心|从一个字符串S构造另一个字符串T，从S的开头或结尾删除一个字符，放在T的尾部。目标：T的字典序尽可能小。|
+|||
 
diff --git a/注意事项.md b/注意事项.md
@@ -3,7 +3,10 @@
 [TOC]
 
 ## 解题思路
-1. 字典序比较类的问题常常用得上贪心法
+1. 字典序比较类的问题常常用得上贪心法。
+2. 深度优先搜索隐式地利用了栈进行计算，占用的内存空间与最大的递归深度有关；而宽度优先搜索则利用了队列，通常需要与状态数成正比的内存空间。一般认为深度优先搜索更省内存
+3. 宽度优先搜索按照距开始状态由近及远的顺序搜索，因此可以很容易的用来求最短路径、最少操作之类的问题。适用于宽度优先搜索的求最短路径，应该具有单位路径长度，或者说求的是最短步数，如果路径带权重，要用Dijkstra等算法。
+4. 
 
 
 
@@ -37,6 +40,7 @@
 
 1. C中没有整除运算，但是可以通过将结果赋值给整数型变量，或强制转换，得到整数结果，例如 int c = 5.6 / 2 ，结果为2，说明浮点数转换到整数是截断，不会舍入。
 2. 数组也可以用sort函数排序，区间为左闭右开。
-3. 模除运算得到的是0到模数-1，不会出现-1，因此不能写i % 80 == -1这种语句
-4. 字符串逆序，有多种办法，可以借用rbegin和rend迭代器，见项目**字典序最小问题_POJ3617-贪心**
-5. string类型字符串可以直接用<比较大小
+3. 模除运算得到的是0到模数-1，不会出现-1，因此不能写i % 80 == -1这种语句。
+4. 字符串逆序，有多种办法，可以借用rbegin和rend迭代器，见项目**字典序最小问题_POJ3617-贪心**。
+5. string类型字符串可以直接用<比较大小。
+6. 有很多地方需要设置一个数为INF，但是不把INF当做例外，而是也参与普通运算的情况也很常见，这种情况下如果INF设置的过大会带来溢出的危险，一般会将INF设置为放大2~4倍也不会溢出的大小。
diff --git a/知识点.md b/知识点.md
@@ -74,3 +74,58 @@ str4  0987654321
 str2  0987654321
 ```
 
+其中reverse函数会将str2逆序，直接打印可以看到，不需要特别的头文件。
+
+
+
+
+
+### stack
+
+##### stack::push()
+
+栈顶放入元素
+
+##### stack::pop()
+
+移除栈顶元素
+
+##### stack::top()
+
+访问栈顶元素
+
+
+
+### queue
+
+##### queue::push()
+
+队列顶部放入元素
+
+##### queue::pop()
+
+移除队列底部的元素
+
+##### queue::front()
+
+访问队列底部的元素
+
+
+
+
+
+### next_permutation函数
+
+
+
+
+
+### 内存
+
+##### 栈内存
+
+栈内存在程序启动时统一分配，此后不能扩大，由于这一区域有上限，所以函数的递归深度也有上限。
+
+##### 堆内存
+
+如果需要申请巨大的数组，将其放置在堆内存上可以减少栈溢出的风险，通常只需定义满足最大需要的数列的大小。