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<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int father[];
int findRoot(int x){
int temp = x;
while(x != father[x]){
x = father[x];
}
int temp2;
while(temp != x){
temp2 = father[temp];
father[temp] = x;
temp = temp2;
}
return x;
}
void union_(int a, int b){
int fa = findRoot(a);
int fb = findRoot(b);
if(fa == fb)
return;
father[fb] = fa;
}
int main(){
int N, K, cnt = -;
for(int i = ; i < ; i++){
father[i] = i;
}
scanf("%d", &N);
for(int i = ; i < N; i++){
int b1, b2, maxb;
scanf("%d%d", &K, &b1);
cnt = max(b1, cnt);
for(int j = ; j < K; j++){
scanf("%d", &b2);
union_(b1, b2);
maxb = max(b1, b2);
cnt = max(maxb, cnt);
}
}
int tree = ;
for(int i = ; i <= cnt; i++){
if(father[i] == i)
tree++;
}
printf("%d %d\n", tree, cnt);
int Q;
scanf("%d", &Q);
for(int i = ; i < Q; i++){
int q1, q2;
scanf("%d%d", &q1, &q2);
if(findRoot(q1) == findRoot(q2))
printf("Yes\n");
else printf("No\n");
}
cin >> N;
return ;
}

1、因为birds并不是连续出现的,需要用set来存储bird的编号。

A1118. Birds in Forest的更多相关文章

  1. PAT A1118 Birds in Forest (25 分)——并查集

    Some scientists took pictures of thousands of birds in a forest. Assume that all the birds appear in ...

  2. PAT甲级——A1118 Birds in Forest【25】

    Some scientists took pictures of thousands of birds in a forest. Assume that all the birds appear in ...

  3. PAT_A1118#Birds in Forest

    Source: PAT A1118 Birds in Forest (25 分) Description: Some scientists took pictures of thousands of ...

  4. 1118 Birds in Forest (25 分)

    1118 Birds in Forest (25 分) Some scientists took pictures of thousands of birds in a forest. Assume ...

  5. PAT1118:Birds in Forest

    1118. Birds in Forest (25) 时间限制 150 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Some ...

  6. [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 t ...

  7. 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 ...

  8. 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  ...

  9. 1118 Birds in Forest (25 分)

    Some scientists took pictures of thousands of birds in a forest. Assume that all the birds appear in ...

随机推荐

  1. checkbox保存和赋值

    //货物信息中的表格内容 $.each(trG.find('td input,td select'),function(i,inp){ if($(inp).attr('type')=='checkbo ...

  2. Linux基础学习笔记2-文件管理和重定向

    本节内容 1)文件系统结构元素 2)创建和查看文件 3)复制.转移和删除文件 4)软和硬链接 5)三种I/O设备 6)把I/O重定向至文件 7)使用管道 文件系统和结构 文件系统 文件和目录被组织成一 ...

  3. linux php7 安装redis扩展

    1,下载redis扩展地址:https://pecl.php.net/package/redis 选择你需要的版本 上传redis-3.1.3.tar.gz到/usr/local/src目录 cd / ...

  4. 百度云虚拟主机配置 Thinkphp5.1

    材料 服务器:百度云虚拟主机(nginx+php7.0+linux) Thinkphp 5.1 问题 百度云默认目录为/webroot,但是我们的需求是将项目存放到/webroot/public下面. ...

  5. size_t的使用

    size_t的取值range是目标平台下最大可能的数组尺寸 典型的例子:x64平台下size_t是8位,而x32平台下是4位: int在两个平台下均为4位 所以在使用的时候一定要配置好对应的平台,否则 ...

  6. c++ 实现拓扑排序

    要简洁大方地实现拓扑排序,首先要了解两个标准模板 std::queue 和 std::vector 1 queue 添加头文件 #include<queue> 定义一个int类型的队列 q ...

  7. kubernetes 编排详解 挂载

    ##kube挂载本地磁盘apiVersion: v1 kind: Pod metadata: name: redis spec: containers: - name: redis image: re ...

  8. [Codeforces1132G]Greedy Subsequences——线段树+单调栈

    题目链接: Codeforces1132G 题目大意:给定一个序列$a$,定义它的最长贪心严格上升子序列为$b$满足若$a_{i}$在$b$中则$a_{i}$之后第一个比它大的也在$b$中.给出一个数 ...

  9. vim复制粘帖代码

    参照:https://blog.csdn.net/sodawaterer/article/details/61918370

  10. 【XSY1551】往事 广义后缀数组 线段树合并

    题目大意 给你一颗trie树,令\(s_i\)为点\(i\)到根的路径上的字符组成的字符串.求\(max_{u\neq v}(LCP(s_u,s_v)+LCS(s_u,s_v))\) \(LCP=\) ...