问题描述: 有个大小为N*M的园子,雨后积起了水.八连通的积水被认为是连接在一起的.求出园子里总共有多少水洼. N, M <= 100 输入例: : 问题分析: 八连通即:上.左上.左,左下,下,右下,右,右上. 这道题可以用深入优先搜索(DFS)的思想. 1. 寻找是水洼的点.如果找到,标记此点已经记过. 循环此点的八连通,如果是水,递归循环 2. 寻找的次数即为水洼数. 代码: # include <iostream> # include <fstream> using…

有一个大小为N*M的园子,雨后积起了水,八连通的积水被认为是链接在一起的求出园子里一共有多少水洼? *** *W* *** /** *进行深度优先搜索,从第一个W开始,将八个方向可以到达的 W修改为 . *每次进行深度优先搜索的时候就将链接的水坑换成了. *进行的深度优先搜索的次数就是水坑数 */ #include<stdio.h> #include<string.h> ; int N,M; char filed[MAX][MAX]; //园子的构造 //现在的位置(x,y) vo…

Lake Counting(POJ No.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')…

题目链接POJ NO.2386 解题思路: 这个也是一个dfs 的应用,在书上的例子,因为书上的代码并不全,基本都是函数分块来写,通过这个题目也规范了代码,以后能用函数的就都用函数来实现吧.采用深度优先搜索,从任意的w开始,不断把邻接的部分用'.'代替,1次DFS后与初始这个w连接的所有w就全都被替换成'.',因此直到图中不再存在W为止,总共进行DFS的次数就是答案.8个方向对应8个状态转移,每个格子作为DFS的参数最多调用一次,因此时间复杂度为O(8nm)=O(nm). AC 代码: #inc…

题目链接:http://poj.org/problem?id=2386 分析:八联通的则为水洼,我们则需遍历一个单位附近的八个单位并将它们都改成'.',但附近单位可能仍连接着有'W'的区域,这种情况下我们应该用dfs遍历到尽头,并将这些联通的W全部改掉,1此dfs后与初始的这个W连接的所有W都替换成了'.',因此直到图中不再存在W为止,最后执行dfs的次数即为水洼的个数. #include <iostream> using namespace std; int n,m; ][]; void d…

Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20003   Accepted: 10063 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 <= 10…

题目描述: 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 ('.'). F…

题目描述:有一个大小为N*M的园子,八连通的积水被认为是连接在一起的.请求出园子里总共有多少水洼?(八连通指的是下图中相对w的*部分) *** *w* *** 限制条件 N,M<=100 样例 输入: N=10,M=12 园子如下图('w'表示积水,'.'表示没有积水) w . . . . . . . . ww . . www . . . . www . . . . ww . . . ww . . . . . . . . . . . ww . . . . . . . . . . . w . .…

[B007]Lake Counting[难度B]—————————————————————————————————————————— [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) square…

数蚂蚁 题目大意:一只牛想数蚂蚁,蚂蚁分成很多组,每个组里面有很多只蚂蚁,现在问你有多少种组合方式 (说白了就是问1,1,1,...,2...,3...,4...)这些东西有多少种排列组合方式 这一道题我一开始想着去用矩阵乘法去做了,结果怎么想怎么不对,后来想着,如果对1,2,3,这些看成背包会怎么样呢?最后结果就压在最后一个背包就可以了 这么一想就懂了,其实就是要你找到递推关系,直接画一个矩阵拉几个箭头就很容易地看出来,对于一个矩阵,dp[i][j]等于dp[i-1][k] j-f[i]<=k…

DFS入门的一道经典题目:LakeCounting 用栈或队列来实现: #include<cstdio> #include<stdlib.h> #include<iostream> #include<stack> using namespace std; int n,m; int pla[10][3]={{1,0},{1,1},{1,-1},{-1,-1},{-1,0},{-1,1},{0,-1},{0,1}};//对坐标进行移动的向量 struct pla…

Lake Counting(POJ No.2386) 有一个大小为N*M的园子,雨后积起了水.八连通的积水被认为是在一起的.请求出园子里共有多少个水洼?(八连通是指下图中相对w的*部分) * * * *w* * * * 限制条件 N,M ≤ 100 思路: 先遍历整个园子,从任意的w开始,把周围所有的w都变成 '.' ,然后遍历完成之后 res 加1 dfs中的操作就是先把当前位置改为'.',然后再把周围上下左右邻近的8个位置全部遍历一遍找w,找到w就从那个w的位置再开始周围8个位置的寻找,一直…

POJ 1852 Ants POJ 2386 Lake Counting POJ 1979 Red and Black AOJ 0118 Property Distribution AOJ 0333 Ball POJ 3009 Curling 2.0 AOJ 0558 Cheese POJ 3669 Meteor Shower AOJ 0121 Seven Puzzle POJ 2718 Smallest Difference POJ 3187 Backward Digit Sums POJ 3…

A - Lake Counting POJ - 2386 最最最最最基础的dfs 挂这道题为了提高AC率(糖水不等式 B - Paint it really, really dark gray CodeForces - 717E dfs 待会写题解 C - New Year Transportation CodeForces - 500A 简单的模拟 D - Binary Tree Traversals HDU - 1710 给树的先序中序输出后序 贴下代码 #include <algorith…

地址 https://leetcode-cn.com/problems/number-of-enclaves/ 给出一个二维数组 A,每个单元格为 0(代表海)或 1(代表陆地). 移动是指在陆地上从一个地方走到另一个地方(朝四个方向之一)或离开网格的边界. 返回网格中无法在任意次数的移动中离开网格边界的陆地单元格的数量. 示例 : 输入:[[,,,],[,,,],[,,,],[,,,]] 输出: 解释: 有三个 被 包围.一个 没有被包围,因为它在边界上. 示例 : 输入:[[,,,],[,,…

A.Lake Counting(POJ 2386) 题意: 由于最近的降雨,农夫约翰田地的各个地方都有水汇聚,用N x M(1 <= N <= 100; 1 <= M <= 100)的矩形表示.每个方格包含水('W')或干燥土地('.').农夫约翰想弄清楚他的田地里形成了多少个池塘.池塘是一组相连的正方形,里面有水,其中一个正方形被认为与八个池塘相邻给定农夫约翰的田野图,确定他有多少个池塘. 思路: 染色法,移动可用(-1,0)类似数组,也可以用for循环,注意循环时递归可能会比在…

ID name status one word  POJ 5437 Alisha’s Party 赛后AC. 优先队列,模拟.对时间t排序 POJ 5438 Ponds 赛后AC 循环链表 POJ 5439 Aggregated Counting     POJ 5440 Clock Adjusting     POJ 5441 Travel 赛后AC 并查集+离线化.注音可以休息. POJ 5442 Favorite Donut     POJ 5443 The Water Problem 题…

http://poj.org/problem?id=2386 http://acm.hdu.edu.cn/showproblem.php?pid=1241 求有多少个连通子图.复杂度都是O(n*m). #include <cstdio> ][]; int n,m; void dfs(int x,int y) { ;i<=;i++) ;j<=;j++) //循环遍历8个方向 { int xx=x+i,yy=y+j; &&xx<n&&yy>=…

http://poj.org/problem?id=2386 题目大意: 有一个大小为N*M的园子,雨后积起了水.八连通的积水被认为是连接在一起的.请求出院子里共有多少水洼? 思路: 水题~直接DFS,DFS过程把途中表示水洼的W改为'.',看DFS了几次即可. #include<cstdio> #include<cstring> const int MAXN=100+10; char map[MAXN][MAXN]; int n,m; void dfs(int x,int y)…

链接:http://poj.org/problem?id=2386 题解 #include<cstdio> #include<stack> using namespace std; ,MAX_N=; char a[MAX_N][MAX_M]; int N,M; //现在位置 (x,y) void dfs(int x,int y){ a[x][y]='.'; //将现在所在位置替换为'.',即旱地 ;dx<=;dx++){ //循环遍历连通的8个方向:上.下.左.右.左上.左下…

地址 http://poj.org/problem?id=2386 <挑战程序设计竞赛>习题 题目描述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 squa…

Lake Counting Time Limit: 1000MS     Memory Limit: 65536K Total Submissions: 17917     Accepted: 9069 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 <=…

Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 28966   Accepted: 14505 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 <= 10…

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…

Lake Counting Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 48370 Accepted: 23775 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…

Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 40370   Accepted: 20015 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 <= 10…

Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18201   Accepted: 9192 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…

-->Lake Counting 直接上中文了 Descriptions: 由于近日阴雨连天,约翰的农场中中积水汇聚成一个个不同的池塘,农场可以用 N x M (1 <= N <= 100; 1 <= M <= 100) 的正方形来表示.农场中的每个格子可以用'W'或者是'.'来分别代表积水或者土地,约翰想知道他的农场中有多少池塘.池塘的定义:一片相互连通的积水.任何一个正方形格子被认为和与它相邻的8个格子相连. 给你约翰农场的航拍图,确定有多少池塘 Input Line 1…

Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 53301   Accepted: 26062 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 <= 10…

好吧前几天一直没更新博客,主要是更新博客的确是要耗费一点精力 北大教你数水坑 最近更新博客可能就是一点旧的东西和一些水题,主要是最近对汇编感兴趣了嘻嘻嘻 这一题挺简单的,没什么难度,简单深搜 #include <stdio.h> #include <stdlib.h> typedef int Postion; ][]; static int N, M; void DFS(Postion, Postion); int main(void) { ; while (~scanf(&quo…