Codeforces 631C. Report 模拟
Each month Blake gets the report containing main economic indicators of the company "Blake Technologies". There are n commodities produced by the company. For each of them there is exactly one integer in the final report, that denotes corresponding revenue. Before the report gets to Blake, it passes through the hands of m managers. Each of them may reorder the elements in some order. Namely, the i-th manager either sorts first ri numbers in non-descending or non-ascending order and then passes the report to the manager i + 1, or directly to Blake (if this manager has number i = m).
Employees of the "Blake Technologies" are preparing the report right now. You know the initial sequence ai of length n and the description of each manager, that is value ri and his favourite order. You are asked to speed up the process and determine how the final report will look like.
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 200 000) — the number of commodities in the report and the number of managers, respectively.
The second line contains n integers ai (|ai| ≤ 109) — the initial report before it gets to the first manager.
Then follow m lines with the descriptions of the operations managers are going to perform. The i-th of these lines contains two integers ti and ri (, 1 ≤ ri ≤ n), meaning that the i-th manager sorts the first ri numbers either in the non-descending (if ti = 1) or non-ascending (if ti = 2) order.
Print n integers — the final report, which will be passed to Blake by manager number m.
3 1
1 2 3
2 2
2 1 3
4 2
1 2 4 3
2 3
1 2
2 4 1 3
In the first sample, the initial report looked like: 1 2 3. After the first manager the first two numbers were transposed: 2 1 3. The report got to Blake in this form.
In the second sample the original report was like this: 1 2 4 3. After the first manager the report changed to: 4 2 1 3. After the second manager the report changed to: 2 4 1 3. This report was handed over to Blake.
题目连接:http://codeforces.com/contest/631/problem/C
题意:一个长度为n的一位数组a,有m操作,每次操作把前r个元素进行非升序或者非降序进行排列。输出最后数组a。
思路:如果后面的操作的元素个数比前面的元素个数多或者相等,前面的操作就会无效。所以优化之后的操作就是r依次递减的。将a的前max(最大的r)位进行非降序排序Q。前一次操作为ti,ri;后一次操作为tj,rj。那么数组a的后ri-rj就确定了,如果ti==1,取剩下的Q的后ri-rj位作为a,如果ti==2,取剩下的Q的前ri-rj作为a。
代码:(Q用数组模拟,标记Q的头s和尾)
#include<bits/stdc++.h>
using namespace std;
struct reorder
{
int p,t,r;
friend bool operator < (reorder a,reorder b)
{
if(a.r!=b.r) return a.r>b.r;
else return a.p>b.p;
}
} reo[];
int a[];
deque<int>Q; ///双向队列
int main()
{
int i,j,n,m;
scanf("%d%d",&n,&m);
for(i=; i<n; i++) scanf("%d",&a[i]);
for(i=; i<m; i++)
{
scanf("%d%d",&reo[i].t,&reo[i].r);
reo[i].p=i+;
}
sort(reo,reo+m);
int cou=reo[].r,sign=reo[].p,flag=reo[].t;
sort(a,a+cou);
for(i=; i<cou; i++) Q.push_back(a[i]); ///队列尾入
reo[m].t=,reo[m].r=,reo[m].p=m+;
for(i=; i<=m; i++)
{
if(reo[i].p>sign)
{
if(flag==)
{
int gg=cou-reo[i].r;
while(gg--)
{
a[--cou]=Q.back(); ///队列尾元素
Q.pop_back(); ///尾列尾出
}
}
else
{
int gg=cou-reo[i].r;
while(gg--)
{
a[--cou]=Q.front(); ///队列头元素
Q.pop_front(); ///队列头出
}
}
sign=reo[i].p;
flag=reo[i].t;
}
}
for(i=; i<n; i++) cout<<a[i]<<" ";
cout<<endl;
return ;
}
数组模拟
关于运算符重载:传送门
关于STL:传送门
Codeforces 631C. Report 模拟的更多相关文章
- Codeforces 631C Report【其他】
题意: 给定序列,将前a个数进行逆序或正序排列,多次操作后,求最终得到的序列. 分析: 仔细分析可以想到j<i,且rj小于ri的操作是没有意义的,对于每个i把类似j的操作删去(这里可以用mult ...
- codeforces 631C. Report
题目链接 按题目给出的r, 维护一个递减的数列,然后在末尾补一个0. 比如样例给出的 4 21 2 4 32 31 2 递减的数列就是3 2 0, 操作的时候, 先变[3, 2), 然后变[2, 0) ...
- codeforces 631C C. Report
C. Report time limit per test 2 seconds memory limit per test 256 megabytes input standard input out ...
- Report CodeForces - 631C (栈)
题目链接 题目大意:给定序列, 给定若干操作, 每次操作将$[1,r]$元素升序或降序排列, 求操作完序列 首先可以发现对最后结果有影响的序列$r$一定非增, 并且是升序降序交替的 可以用单调栈维护这 ...
- Codeforces 389B(十字模拟)
Fox and Cross Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u Submi ...
- codeforces 591B Rebranding (模拟)
Rebranding Problem Description The name of one small but proud corporation consists of n lowercase E ...
- Codeforces 626B Cards(模拟+规律)
B. Cards time limit per test:2 seconds memory limit per test:256 megabytes input:standard input outp ...
- Codeforces 679B. Barnicle 模拟
B. Barnicle time limit per test: 1 second memory limit per test :256 megabytes input: standard input ...
- CodeForces 382C【模拟】
活生生打成了大模拟... #include <bits/stdc++.h> using namespace std; typedef long long LL; typedef unsig ...
随机推荐
- PyCharm License Activation激活码失效问题的解决方法
自己的小Python项目好几天没有写了,今天打开PyCharm准备继续写,突然发现之前的激活码被取消不能用了,本来激情满满的准备干活啦!之前搜的这个激活码本来说的是可以用到18年1月份的呢,哎…心情是 ...
- 在oracle下如何创建database link全面总结
物理上存放于网络的多个ORACLE数据库,逻辑上可以看成一个单一的大型数据库,用户可以通过网络对异地数据库中的数据进行存取,而服务器之间的协同处理对于工作站用户及应用程序而言是完全透明的,开发人员无需 ...
- shutil模块---文件,文件夹复制、删除、压缩等处理
shutil模块:高级的文件,文件夹,压缩包处理 拷贝内容 # shutil.copyfileobj(open('example.ini','r'),open('example.new','w')) ...
- ANA网络分析
ANN网络分析 Mnist手写数字识别 Mnist数据集可以从官网下载,网址: http://yann.lecun.com/exdb/mnist/ 下载下来的数据集被分成两部分:55000行的训练数据 ...
- Nginx-ingress-controller部署
参考官网https://kubernetes.github.io/ingress-nginx/ 部署pod:nginx-ingress-controller/nginx-default-backend ...
- 1.urlencoder和urldecoder的使用
今天传url的时候乱码了.先说情形,url中有searchText=中文的情形,后台new String(searchText.getBytes(ISO-8859-1),"gbk" ...
- java.lang.ClassCastException: java.util.Arrays$ArrayList cannot be cast to java.util.ArrayList
String[] 转换成 ArrayList 报的错. String[] str = {"A","B"}; ArrayList<String> li ...
- Redis基本操作-list
Redis的5种数据结构:string.list.hash.set和zset; Redis 所有的数据结构都是以唯一的 key 字符串作为名称,然后通过这个唯一 key 值来获取相应的 value 数 ...
- How to Pronounce BEAUTIFUL
How to Pronounce BEAUTIFUL Share Tweet Share Tagged With: 3-Syllable Can you say this word beautiful ...
- ML_入门
N-gram 输入法后来提醒nlp自然语言理解一个向量映射到另一个空间,为什么是向量呢?模型其实是向量,一张图片表示成向量,像素表示成rgb ,每一个维度 数的度文本变成向量 one-hot repr ...