hdu 1198 Farm Irrigation(并查集)
题意:
Benny has a spacious farm land to irrigate. The farm land is a rectangle, and is divided into a lot of samll squares. Water pipes are placed in these squares. Different square has a different type of pipe. There are 11 types of pipes, which is marked from A to K, as Figure 1 shows.
Benny has a map of his farm, which is an array of marks denoting the distribution of water pipes over the whole farm. For example, if he has a map
ADC
FJK
IHE
then the water pipes are distributed like
Several wellsprings are found in the center of some squares, so water can flow along the pipes from one square to another. If water flow crosses one square, the whole farm land in this square is irrigated and will have a good harvest in autumn.
Now Benny wants to know at least how many wellsprings should be found to have the whole farm land irrigated. Can you help him?
There are several test cases! In each test case, the first line contains 2 integers M and N, then M lines follow. In each of these lines, there are N characters, in the range of 'A' to 'K', denoting the type of water pipe over the corresponding square. A negative M or N denotes the end of input, else you can assume 1 <= M, N <= 50.
思路:
并查集,看代码。
代码:
struct node{
bool up,down,left,right;
}
a[2505]; int m,n;
char graph[55][55];
int fa[2505]; int findFa(int x){
return fa[x]==x?fa[x]:fa[x]=findFa(fa[x]);
} void Union(int x,int y){
int fx=findFa(x);
int fy=findFa(y);
if(fx!=fy){
fa[fx]=fy;
}
} int main(){ while(scanf("%d%d",&m,&n),m>0||n>0){
rep(i,0,m-1) scanf("%s",graph[i]); rep(i,1,m*n) a[i].up=a[i].down=a[i].left=a[i].right=false; rep(i,0,m-1){
rep(j,0,n-1){
switch(graph[i][j]){
case 'A':a[i*n+j+1].up=true; a[i*n+j+1].left=true; break;
case 'B':a[i*n+j+1].up=true; a[i*n+j+1].right=true; break;
case 'C':a[i*n+j+1].left=true; a[i*n+j+1].down=true; break;
case 'D':a[i*n+j+1].right=true; a[i*n+j+1].down=true; break;
case 'E':a[i*n+j+1].up=true; a[i*n+j+1].down=true; break;
case 'F':a[i*n+j+1].left=true; a[i*n+j+1].right=true; break;
case 'G':a[i*n+j+1].left=true; a[i*n+j+1].right=true; a[i*n+j+1].up=true; break;
case 'H':a[i*n+j+1].up=true; a[i*n+j+1].down=true; a[i*n+j+1].left=true; break;
case 'I':a[i*n+j+1].right=true; a[i*n+j+1].left=true; a[i*n+j+1].down=true; break;
case 'J':a[i*n+j+1].up=true; a[i*n+j+1].down=true; a[i*n+j+1].right=true; break;
case 'K':a[i*n+j+1].up=true; a[i*n+j+1].left=true; a[i*n+j+1].down=true; a[i*n+j+1].right=true; break;
default: break;
}
}
} rep(i,1,m*n) fa[i]=i;
rep(i,1,m){
rep(j,1,n){
int now=(i-1)*n+j;
if(j!=1){
int last1=(i-1)*n+j-1;
if(a[last1].right && a[now].left){
Union(last1,now);
}
}
if(i!=1){
int last2=(i-2)*n+j;
if(a[last2].down && a[now].up){
Union(last2,now);
}
}
}
} map<int,int> mp;
mp.clear();
int ans=0;
rep(i,1,m*n){
int t=findFa(i);
if(mp[t]==0){
++ans;
mp[t]=1;
}
} printf("%d\n",ans); } return 0;
}
hdu 1198 Farm Irrigation(并查集)的更多相关文章
- 杭电OJ——1198 Farm Irrigation (并查集)
畅通工程 Problem Description 某省调查城镇交通状况,得到现有城镇道路统计表,表中列出了每条道路直接连通的城镇.省政府"畅通工程"的目标是使全省任何两个城镇间都可 ...
- hdu 1198 Farm Irrigation(深搜dfs || 并查集)
转载请注明出处:viewmode=contents">http://blog.csdn.net/u012860063?viewmode=contents 题目链接:http://acm ...
- HDU 1198 Farm Irrigation(并查集,自己构造连通条件或者dfs)
Farm Irrigation Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)T ...
- HDU 1198 Farm Irrigation(并查集+位运算)
Farm Irrigation Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other) Tot ...
- HDU 1198 Farm Irrigation(状态压缩+DFS)
题目网址:http://acm.hdu.edu.cn/showproblem.php?pid=1198 题目: Farm Irrigation Time Limit: 2000/1000 MS (Ja ...
- HDU 1198 Farm Irrigation (并检查集合 和 dfs两种实现)
Farm Irrigation Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
- hdu1198 Farm Irrigation 并查集
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1198 简单并查集 分别合并竖直方向和水平方向即可 代码: #include<iostream&g ...
- hdu.1198.Farm Irrigation(dfs +放大建图)
Farm Irrigation Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
- hdu 1198 Farm Irrigation
令人蛋疼的并查集…… 我居然做了大量的枚举,居然过了,我越来越佩服自己了 这个题有些像一个叫做“水管工”的游戏.给你一个m*n的图,每个单位可以有11种选择,然后相邻两个图只有都和对方连接,才判断他们 ...
随机推荐
- 最新版微软视窗(Windows)作业系统下载(2020-08-19)
为了更好的使用WSL(Windows Subsystem For Linux),不得不用最新的windows 10 2004版了,这个版本的WSL已经是第二版了,即WSL2.下面给出下载地址 系统发布 ...
- 深入学习Composer原理(一)
Composer作为PHP的包管理工具,为PHPer们提供了丰富的类库,并且让PHP重焕新生,避免被时代淘汰的悲剧.可以说,Composer和PHP7是现在PHP开发者的标配,如果你还没用过Compo ...
- linux停止进程
$ kill -QUIT 主进程号 例如:kill -QUIT 1708 这种方式相对来说会有一个停止的过程,先将子进程停止掉,然后再停掉主进程. 快速停止Nginx: $ kill -TERM 主进 ...
- django 各项配置基本设置
setting中一些设置例子 mysql数据库连接设置 DATABASES = { 'default': { 'ENGINE': 'django.db.backends.mysql', 'NAME': ...
- django ORM教程(转载)
Django中ORM介绍和字段及字段参数 Object Relational Mapping(ORM) ORM介绍 ORM概念 对象关系映射(Object Relational Mapping,简 ...
- web带宽估算方法
每个连接约占用10Kb的带宽,以3万总用户数和10%的在线率计算,并按照10%的冗余率,服务器总带宽=每秒总连接数*10Kbps /(1-冗余率)/1024. 带宽占用(Mbps)=30000*10% ...
- Spirit带你彻底搞懂JS的6种继承方案
JavaScript中实现继承的6种方案 01-原型链的继承方案 function Person(){ this.name="czx"; } function Student(){ ...
- 通过Git在本地局域网中的两台电脑间同步代码
0.前言 一般情况下同步代码可以通过在GitHub/GitLab等网站新建远程仓库,所有机器都向仓库推送或者从仓库下拉更新. 上述过程步骤也不算复杂,不过有时候我们考虑到仓库的安全性等因素,只想在局域 ...
- Parameter index out of range(1 > number of parameters, which is 0)参数索引超出范围
今天在写项目的过程中,有一个模块是做多选删除操作,通过servlet获得多选框的value组,然后执行sql操作.如下: 1 @RequestMapping( "/delteCouse.do ...
- Bloom Filter算法
Bloom Filter算法详解 什么是布隆过滤器 布隆过滤器(Bloom Filter)是 1970 年由布隆提出的.它实际上是一个很长的二进制向量和一系列随机映射函数 (下面详细说),实际上你也可 ...