CF 843 A. Sorting by Subsequences
You are given a sequence a1, a2, ..., an consisting of different integers. It is required to split this sequence into the maximum number of subsequences such that after sorting integers in each of them in increasing order, the total sequence also will be sorted in increasing order.
Sorting integers in a subsequence is a process such that the numbers included in a subsequence are ordered in increasing order, and the numbers which are not included in a subsequence don't change their places.
Every element of the sequence must appear in exactly one subsequence.
The first line of input data contains integer n (1 ≤ n ≤ 105) — the length of the sequence.
The second line of input data contains n different integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109) — the elements of the sequence. It is guaranteed that all elements of the sequence are distinct.
In the first line print the maximum number of subsequences k, which the original sequence can be split into while fulfilling the requirements.
In the next k lines print the description of subsequences in the following format: the number of elements in subsequence ci (0 < ci ≤ n), then ci integers l1, l2, ..., lci (1 ≤ lj ≤ n) — indices of these elements in the original sequence.
Indices could be printed in any order. Every index from 1 to n must appear in output exactly once.
If there are several possible answers, print any of them.
6
3 2 1 6 5 4
4
2 1 3
1 2
2 4 6
1 5
6
83 -75 -49 11 37 62
1
6 1 2 3 4 5 6
In the first sample output:
After sorting the first subsequence we will get sequence 1 2 3 6 5 4.
Sorting the second subsequence changes nothing.
After sorting the third subsequence we will get sequence 1 2 3 4 5 6.
Sorting the last subsequence changes nothing.
把每一次交换涉及到的元素放到一个集合中。
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 1044266558
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
int a[],b[],vis[];
int ans,n;
set<int>s;
set<int>::iterator it;
int main()
{
while(scanf("%d",&n)!=EOF)
{
fill(vis,vis+n,);
ans=;
for(int i=;i<n;i++) scanf("%d",&a[i]),b[i]=a[i];
sort(b,b+n);
for(int i=;i<n;i++) a[i]=lower_bound(b,b+n,a[i])-b;
for(int i=;i<n;i++)
{
if(!vis[i])
{
for(int j=a[i];!vis[j];j=a[j]) vis[j]=;
ans++;
}
}
printf("%d\n",ans);
for(int i=;i<n;i++)
{
if(vis[i])
{
s.clear();
for(int j=a[i];vis[j];j=a[j]) s.insert(j+),vis[j]=;
int pos=s.size();
printf("%d",pos);
for(it=s.begin();it!=s.end();it++)
printf(" %d",*it);
printf("\n");
}
}
}
return ;
}
CF 843 A. Sorting by Subsequences的更多相关文章
- cf 843 A Sorting by Subsequences [建图]
题面: 传送门 思路: 这道题乍一看有点难 但是实际上研究一番以后会发现,对于每一个位置只会有一个数要去那里,还有一个数要离开 那么只要把每个数和他将要去的那个位置连起来,构成了一个每个点只有一个入边 ...
- cf 843 D Dynamic Shortest Path [最短路+bfs]
题面: 传送门 思路: 真·动态最短路 但是因为每次只加1 所以可以每一次修改操作的时候使用距离分层的bfs,在O(n)的时间内解决修改 这里要用到一个小技巧: 把每条边(u,v)的边权表示为dis[ ...
- cf 843 B Interactive LowerBound [随机化]
题面: 传送门 思路: 这是一道交互题 比赛的时候我看到了直接跳过了...... 后来后面的题目卡住了就回来看这道题,发现其实比较水 实际上,从整个序列里面随机选1000个数出来询问,然后从里面找出比 ...
- 【AIM Tech Round 4 (Div. 2) C】Sorting by Subsequences
[链接]http://codeforces.com/contest/844/problem/C [题意] 水题,没有记录意义 [题解] 排序之后,记录每个数字原来在哪里就好. 可以形成环的. 环的个数 ...
- AIM Tech Round 4 (Div. 2)ABCD
A. Diversity time limit per test 1 second memory limit per test 256 megabytes input standard input o ...
- AIM Tech Round 4 (Div. 2)(A,暴力,B,组合数,C,STL+排序)
A. Diversity time limit per test:1 second memory limit per test:256 megabytes input:standard input o ...
- 【Codeforces AIM Tech Round 4 (Div. 2) C】
·将排序限制于子序列中,又可以说明什么呢? C. Sorting by Subsequences ·英文题,述大意: 输入一个长度为n的无重复元素的序列{a1,a2……an}(1<= ...
- CF 689D - Friends and Subsequences
689D - Friends and Subsequences 题意: 大致跟之前题目一样,用ST表维护a[]区间max,b[]区间min,找出多少对(l,r)使得maxa(l,r) == minb( ...
- CF#335 Sorting Railway Cars
Sorting Railway Cars time limit per test 2 seconds memory limit per test 256 megabytes input standar ...
随机推荐
- Win10 + YOLOv3训练VOC数据集-----How to train Pascal VOC Data
How to train (Pascal VOC Data): Download pre-trained weights for the convolutional layers (154 MB): ...
- 紫书 习题 10-3 UVa 1643(计算几何 叉乘)
直观感觉对角线重合的时候面积最大 然后可以根据方程和割补算出阴影部分的面积 注意知道两点坐标,可以求出与原点形成的三角形的面积 用叉乘,叉乘的几何意义以这两个向量为边的平行四边形的面积 所以用叉乘除以 ...
- 最全面的AndroidStudio配置指南总结-包括护眼模式
使用AndroidStudio开发APP已有半年多的时间了,从刚开始的不习惯到慢慢适应再到逐渐喜欢上AndroidStudio,中间的过程颇有一番曲折,现在把自己对AndroidStudio的配置心得 ...
- static_cast 与 dynamic_cast
- HDU 4183Pahom on Water(网络流之最大流)
题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=4183 这题题目意思非常难看懂..我看了好长时间也没看懂..终于是从网上找的翻译. .我就在这翻译一下吧 ...
- JavaScript语言基础3
JavaScript能够处理一些来自于现实世界的数据类型.比如:数字和文本. 同一时候JavaScript中也包括了一些具 有抽象性质的数据类型.比如对象数据类型. JavaScript它是一种弱类 ...
- 解决Struts中文乱码问题总结
在进行struts开发的过程中.总也是出现非常多的乱码问题.但归根究竟,也仅仅是下面三种情况: ㈠页面显示中文乱码 ㈡传递參数中文乱码 ㈢国际化资源文件乱码 以下就这三中情况介绍怎么在详细项目 ...
- 巧用FPGA中资源
随着FPGA的广泛应用,所含的资源也越来越丰富,从基本的逻辑单元.DSP资源和RAM块,甚至CPU硬核都能集成在一块芯片中.在做FPGA设计时,如果针对FPGA中资源进行HDL代码编写,对设计的资源利 ...
- netflix turbine概述
1.turbine是什么?它的作用是什么? Turbine is a tool for aggregating streams of Server-Sent Event (SSE) JSON data ...
- Python(六) Python 函数
一.认识函数 help(方法名字) help(round) 1.功能性 2.隐藏细节 3.避免编写重复的代码 4.组织代码 自定义函数 二.函数的定义及运行特点 # 递归 def sum_num(n ...