题目大意:有N*M的矩阵稻田,'W'表示有积水的地方, '.'表示是干旱的地方,问稻田内一共有多少块积水,根据样例很容易得出,积水是8个方向任一方向相连即可. 题目大意:有N*M的矩阵稻田,'W'表示有积水的地方, '.'表示是干旱的地方,问稻田内一共有多少块积水,根据样例很容易得出,积水是8个方向任一方向相连即可.  题目大意:有N*M的矩阵稻田,'W'表示有积水的地方, '.'表示是干旱的地方,问稻田内一共有多少块积水,根据样例很容易得出,积水是8个方向任一方向相连即可.  题目大意:有N*…

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…

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…

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…

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: 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 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 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: 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…

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)…

题意:给定一个n*m的矩阵,让你判断有多少个连通块. 析:用DFS搜一下即可. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring&…

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…

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…

简单的深度搜索就能够了,看见有人说什么使用并查集,那简直是大算法小用了. 由于能够深搜而不用回溯.故此效率就是O(N*M)了. 技巧就是添加一个标志P,每次搜索到池塘,即有W字母,那么就觉得搜索到一个池塘了,P值为真. 搜索过的池塘不要反复搜索,故此,每次走过的池塘都改成其它字母.如'@',或者'#',随便一个都能够. 然后8个方向搜索. #include <stdio.h> #include <vector> #include <string.h> #include…

题意:给出一个矩形,问有多少块连通的W 当找到W的时候,进行广搜,然后将搜过的W变成点,直到不能再搜,进行下一次广搜,最后搜的次数即为水塘的个数 看的PPT里面讲的是种子填充法. 种子填充算法: 从多边形区域的一个内点开始,由内向外用给定的颜色画点直到边界为止.如果边界是以一种颜色指定的,则种子填充算法可逐个像素地处理直到遇到边界颜色为止 对于这一题: 先枚举矩阵中的每一个元素,当元素为W的时候,对它进行种子填充(BFS) 种子填充过程: 1)将八个方向的状态分别加进队列 2)如果元素为W,将其…

来之不易的2017第一发ac http://poj.org/problem?id=2386 Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 31474   Accepted: 15724 Description Due to recent rains, water has pooled in various places in Farmer John's field, which is repr…

-->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…

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…

Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 49414   Accepted: 24273 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…

http://blog.csdn.net/c20182030/article/details/52327948 1388:Lake Counting 总时间限制:   1000ms   内存限制:   65536kB 描述 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 <= 1…

Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20782   Accepted: 10473 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…

很水的DFS. 为什么放上来主要是为了让自己的博客有一道DFS题解,,, #include<bits/stdc++.h> using namespace std; ][],ans,flag; char sx; void dfs(int x,int y){ ){ flag=; a[x][y]=; dfs(x-,y); dfs(x+,y); dfs(x,y-); dfs(x,y+); dfs(x-,y-); dfs(x+,y-); dfs(x-,y+); dfs(x+,y+); } } int m…

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…

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…

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 . .…

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')…