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…

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)…

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: 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 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>=…

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…

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&…

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*…

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…

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…

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…

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: 16802   Accepted: 8523 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…

棋盘问题 Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 44012   Accepted: 21375 Description 在一个给定形状的棋盘(形状可能是不规则的)上面摆放棋子,棋子没有区别.要求摆放时任意的两个棋子不能放在棋盘中的同一行或者同一列,请编程求解对于给定形状和大小的棋盘,摆放k个棋子的所有可行的摆放方案C. Input 输入含有多组测试数据.每组数据的第一行是两个正整数,n k,用一个空格隔开,表示了…

传送门:http://poj.org/problem?id=1287 题意:给出n个点 m条边 ,求最小生成树的权 思路:最小生树的模板题,直接跑一遍kruskal即可 代码: #include<iostream> #include<cstdio> #include<algorithm> #include<string.h> using namespace std; const int maxn = 5005; struct node { int u; in…

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…

N*M的园子,雨后积起了水.八连通的积水背认为是连接在一起的.请求出园子里总共有多少水洼? dfs(Depth-First  Search)  八个方向的简单搜索.... 深度优先搜索从最开始的状态出发,遍历所有可以到达的状态.由此可以对所有的状态进行操作,或者列举出所有的状态. int N,M; ][]; void dfs(int x,int y) { field[x][y]=='.'; //将现在所在位置替换 ; dx<=; dx++){ ; dy<=; dy++){ //向x方向移动dx…

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…

Time limit1000 ms Memory limit65536 kB 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')…

poj 1251  && hdu 1301 Sample Input 9 //n 结点数A 2 B 12 I 25B 3 C 10 H 40 I 8C 2 D 18 G 55D 1 E 44E 2 F 60 G 38F 0G 1 H 35H 1 I 353A 2 B 10 C 40B 1 C 200Sample Output 21630 prim算法 # include <iostream> # include <cstdio> # include <cstr…

题目链接:http://poj.org/problem?id=1986 Description Farmer John's cows refused to run in his marathon since he chose a path much too long for their leisurely lifestyle. He therefore wants to find a path of a more reasonable length. The input to this prob…

题目链接:http://poj.org/problem?id=1269 Time Limit: 1000MS Memory Limit: 10000K Description We all know that a pair of distinct points on a plane defines a line and that a pair of lines on a plane will intersect in one of three ways: 1) no intersection b…

http://poj.org/problem?id=2186 首先求出所有的强连通分量,分好块.然后对于每一个强连通分量,都标记下他们的出度.那么只有出度是0 的块才有可能是答案,为什么呢?因为既然你有了出度,那么就是指向了另外一个块,那么你就不能指望那个块也指向你了,因为这样会形成环,所以肯定有一个块的cow不支持你,所以你这个块就不会是popular cow 那么就有两种情况, ①.出度是0的块的个数是1个,那么就是唯一的答案. ②.出度是0的快的个数有多个,那么答案是0, 因为至少也有一个…

原题传送:http://poj.org/problem?id=3461 Oulipo Time Limit: 1000MS Memory Limit: 65536K Description The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from th…