[并查集] 1118. Birds in Forest (25)
1118. Birds in Forest (25)
Some scientists took pictures of thousands of birds in a forest. Assume that all the birds appear in the same picture belong to the same tree. You are supposed to help the scientists to count the maximum number of trees in the forest, and for any pair of birds, tell if they are on the same tree.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive number N (<= 104) which is the number of pictures. Then N lines follow, each describes a picture in the format:
K B1 B2 ... BK
where K is the number of birds in this picture, and Bi's are the indices of birds. It is guaranteed that the birds in all the pictures are numbered continuously from 1 to some number that is no more than 104.
After the pictures there is a positive number Q (<= 104) which is the number of queries. Then Q lines follow, each contains the indices of two birds.
Output Specification:
For each test case, first output in a line the maximum possible number of trees and the number of birds. Then for each query, print in a line "Yes" if the two birds belong to the same tree, or "No" if not.
Sample Input:
4
3 10 1 2
2 3 4
4 1 5 7 8
3 9 6 4
2
10 5
3 7
Sample Output:
2 10
Yes
No
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std; const int maxn=1e4+; int father[maxn];
int num[maxn]={};
int flag[maxn]={}; int findFather(int x)
{
int a=x;
while(x!=father[x])
{
x=father[x];
}
//路径压缩
while(a!=father[a])
{
int z=a;
a=father[a];
father[z]=x;
}
return x;
} void uf(int a,int b)
{
int fa=findFather(a);
int fb=findFather(b);
if(fa!=fb)
{
father[fa]=fb;
num[fa]+=num[fb];
}
} int max_num=;
int ans=; int main()
{
for(int i=;i<maxn;i++) father[i]=i;
fill(num,num+maxn,);
int n;
scanf("%d",&n);
for(int i=;i<n;i++)
{
int k,first;
scanf("%d %d",&k,&first);
flag[first]=;
max_num=first>max_num?first:max_num;
for(int j=;j<k;j++)
{
int a;
scanf("%d",&a);
flag[a]=;
max_num=max_num>a?max_num:a;
uf(a,first);
}
}
int cnt=;
for(int i=;i<=max_num;i++)
{
//printf(" %d",father[i]);
if(father[i]==i) cnt++;
}
for(int i=;i<maxn;i++) ans+=flag[i];
printf("%d %d\n",cnt,ans);
int q;
scanf("%d",&q);
for(int i=;i<q;i++)
{
int u,v;
scanf("%d%d",&u,&v);
if(findFather(u)==findFather(v))
{
printf("Yes\n");
}
else
{
printf("No\n");
}
}
return ;
}
[并查集] 1118. Birds in Forest (25)的更多相关文章
- PAT A 1118. Birds in Forest (25)【并查集】
并查集合并 #include<iostream> using namespace std; const int MAX = 10010; int father[MAX],root[MAX] ...
- PAT题解-1118. Birds in Forest (25)-(并查集模板题)
如题... #include <iostream> #include <cstdio> #include <algorithm> #include <stri ...
- 1118. Birds in Forest (25)
Some scientists took pictures of thousands of birds in a forest. Assume that all the birds appear in ...
- 【PAT甲级】1118 Birds in Forest (25分)(并查集)
题意: 输入一个正整数N(<=10000),接着输入N行数字每行包括一个正整数K和K个正整数,表示这K只鸟是同一棵树上的.输出最多可能有几棵树以及一共有多少只鸟.接着输入一个正整数Q,接着输入Q ...
- PAT甲级——1118 Birds in Forest (并查集)
此文章 同步发布在CSDN:https://blog.csdn.net/weixin_44385565/article/details/89819984 1118 Birds in Forest ...
- 1118 Birds in Forest (25 分)
1118 Birds in Forest (25 分) Some scientists took pictures of thousands of birds in a forest. Assume ...
- PAT 1118 Birds in Forest [一般]
1118 Birds in Forest (25 分) Some scientists took pictures of thousands of birds in a forest. Assume ...
- PAT 1118 Birds in Forest
Some scientists took pictures of thousands of birds in a forest. Assume that all the birds appear in ...
- 1118 Birds in Forest
题意: 思路:并查集模板题. 代码: #include <cstdio> #include <algorithm> using namespace std; ; int fat ...
随机推荐
- LZO压缩算法64位崩溃问题
*** vs2013 64位调用LZO算法失败,原因: vs2013 long 类型4位 指针为8位. 解决: 将static lzo_bool basic_ptr_check(void)函数中,指针 ...
- 2017战略No.1:坚定不移地走全产业链发展路线
编者按:2016年9月9日,首次公开表达"我想走全产业链发展路线"的想法. 这几个月,认真思考了下这个决定背后的原因.目的和价值. 付出常人5倍以上的努力,先抓住"技术研 ...
- JavaScript总结(三)
如何执行代码语句? 使用函数,函数是一组可以随时随地运行的语句,它们是JavaScript的核心.函数是由关键字function.函数名加一组参数以及置于括号中要执行的代码声明的.语法如下: Func ...
- exLucas学习笔记
exLucas学习笔记 Tags:数学 写下抛硬币和超能粒子炮改 洛谷模板代码如下 #include<iostream> #define ll long long using namesp ...
- SpringMVC初写(三)Controller的生命周期
Spring框架默认创建的对象的方式是单例,所以业务控制器Controller也是一个单例对象 由此可证明,无论是同一次请求还是同一次会话和不同请求它的对象都是相同的 然而由于对象是单例的,随之而来的 ...
- log4j配置单独日志文件输出
log4j.logger.batteryHistory=ERROR,BD log4j.appender.BD=org.apache.log4j.FileAppender log4j.appender. ...
- pathon之多线程详解
一.线程理论 1.什么是线程 线程指的是一条流水线的工作过程 进程根本就不是一个执行单位,进程其实是一个资源单位--------将资源集合到一起: 一个进程内自带一个线程,线程才是CPU上的执行单位 ...
- Package设计1:选择数据类型、暂存数据和并发
SSIS 设计系列: Package设计1:选择数据类型.暂存数据和并发 Package设计2:增量更新 Package 设计3:数据源的提取和使用暂存 一,数据类型的选择 对于SSIS的数据类型,容 ...
- Win7搭建FTP服务器
“控制面板” -> “程序和功能” -> “打开或关闭Windows 功能”: 1.展开“Internet 信息服务” 2.勾选“Internet Information Services ...
- linux 查询管道过滤,带上标题字段
linux查询过滤, 带上标题字段例: 一个简单的查询 ps -e | grep httpd 上面经过grep 过滤后, 标题没了, 但是为了看上去更方便,有标题字段看起来更方便一些, 那么可以按下面 ...