原博主:https://blog.csdn.net/amovement/article/details/80358962
B. Bus of Characters
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

In the Bus of Characters there are nn rows of seat, each having 22 seats. The width of both seats in the ii-th row is wiwi centimeters. All integers wiwi are distinct.

Initially the bus is empty. On each of 2n2n stops one passenger enters the bus. There are two types of passengers:

  • an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it;
  • an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it.

You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take.

Input

The first line contains a single integer nn (1≤n≤2000001≤n≤200000) — the number of rows in the bus.

The second line contains the sequence of integers w1,w2,…,wnw1,w2,…,wn (1≤wi≤1091≤wi≤109), where wiwi is the width of each of the seats in the ii-th row. It is guaranteed that all wiwi are distinct.

The third line contains a string of length 2n2n, consisting of digits '0' and '1' — the description of the order the passengers enter the bus. If the jj-th character is '0', then the passenger that enters the bus on the jj-th stop is an introvert. If the jj-th character is '1', the the passenger that enters the bus on the jj-th stop is an extrovert. It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal nn), and for each extrovert there always is a suitable row.

Output

Print 2n2n integers — the rows the passengers will take. The order of passengers should be the same as in input.

Examples
input

Copy
2
3 1
0011
output

Copy
2 1 1 2
input

Copy
6
10 8 9 11 13 5
010010011101
output

Copy
6 6 2 3 3 1 4 4 1 2 5 5
Note

In the first example the first passenger (introvert) chooses the row 22, because it has the seats with smallest width. The second passenger (introvert) chooses the row 11, because it is the only empty row now. The third passenger (extrovert) chooses the row 11, because it has exactly one occupied seat and the seat width is the largest among such rows. The fourth passenger (extrovert) chooses the row 22, because it is the only row with an empty place.

一、原题地址

http://codeforces.com/contest/982/problem/B

二、大致题意

一辆公交有n排座位,每排座位有两个椅子。

接着给出每排座位的宽度。

这时有2*n个人要上车,这群人有两种类型。

1、内向的人:会优先选择座位宽度最小的那排就坐,最好旁边是没人的。

2、外向的人:会优先选择座位旁边有人的那排就坐,在此基础上座位选择越大越好。

三、思路

我们可以先把座位的宽度进行排序。

1、当内向的人上车时,我们记录一个pos表示当前内向的人最优的位置,同时将座位的编号存入栈。

2、当外向的人上车时,我们提取栈内最顶端的元素,因为内向的人是按照座位宽度从小到大入座的,所以此时的栈顶元素,必定为有内向的人就坐的宽度最大的位置。同时删除栈顶元素,表示这个位置被使用了。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<set>
#include<map>
#include<stack>
using namespace std;
const int inf = 0x3f3f3f3f;
long long gcd(long long a, long long b) { return a == ? b : gcd(b % a, a); } int n;
struct SEAT
{
int w, id;
}seat[];
bool cmp(SEAT xx, SEAT yy)
{
return xx.w < yy.w;
}
int main()
{
scanf("%d", &n);
for (int i = ; i <= n; i++)
{
scanf("%d", &seat[i].w);
seat[i].id = i;
}
sort(seat + , seat + + n, cmp);
char s[ * ];
scanf("%s", s);
int pos0 = ;
stack<int>sta;
for (int i = ; i < * n; i++)
{
if (s[i] == '')
{
printf("%d ", seat[pos0].id);
sta.push(seat[pos0].id);
pos0++;
}
else if (s[i] == '')
{
printf("%d ", sta.top());
sta.pop();
}
}
getchar();
getchar();
}

Codeforces Round #484 (Div. 2) B. Bus of Characters(STL+贪心)982B的更多相关文章

  1. Codeforces Codeforces Round #484 (Div. 2) E. Billiard

    Codeforces Codeforces Round #484 (Div. 2) E. Billiard 题目连接: http://codeforces.com/contest/982/proble ...

  2. Codeforces Codeforces Round #484 (Div. 2) D. Shark

    Codeforces Codeforces Round #484 (Div. 2) D. Shark 题目连接: http://codeforces.com/contest/982/problem/D ...

  3. Codeforces Round #484 (Div. 2)

    题目链接:http://codeforces.com/contest/982 A. Row time limit per test:1 second memory limit per test:256 ...

  4. Codeforces Round #285 (Div. 2) A B C 模拟 stl 拓扑排序

    A. Contest time limit per test 1 second memory limit per test 256 megabytes input standard input out ...

  5. Codeforces Round #436 (Div. 2)C. Bus 模拟

    C. Bus time limit per test: 2 seconds memory limit per test: 256 megabytes input: standard input out ...

  6. Codeforces Round #369 (Div. 2) A. Bus to Udayland 水题

    A. Bus to Udayland 题目连接: http://www.codeforces.com/contest/711/problem/A Description ZS the Coder an ...

  7. Codeforces Round #369 (Div. 2) A. Bus to Udayland (水题)

    Bus to Udayland 题目链接: http://codeforces.com/contest/711/problem/A Description ZS the Coder and Chris ...

  8. Codeforces Round #436 (Div. 2) C. Bus

    http://codeforces.com/contest/864/problem/C 题意: 坐标轴上有x = 0和 x = a两点,汽车从0到a之后掉头返回,从a到0之后又掉头驶向a...从0到a ...

  9. Codeforces Round #491 (Div. 2) E - Bus Number + 反思

    E - Bus Number 最近感觉打CF各种车祸.....感觉要反思一下, 上次读错题,这次想当然地以为18!肯定暴了longlong 而没有去实践, 这个题我看到就感觉是枚举每个数字的个数,但是 ...

随机推荐

  1. CSS开启硬件加速来提高网站性能

    原文永久链接 CSS animations, transforms 以及 transitions 不会自动开启GPU加速,而是由浏览器的缓慢的软件渲染引擎来执行. 那我们怎样才可以切换到GPU模式呢, ...

  2. (转)Linux LVM逻辑卷配置过程详解(创建、扩展、缩减、删除、卸载、快照创建)

    一.预备知识 LVM全称为Logical Volume Manager 逻辑卷管理器,LVM是Linux环境中对磁盘分区进行管理的一种机制,是建立在硬盘和分区之上.文件系统之下的一个逻辑层,可提高磁盘 ...

  3. Java内部类超详细总结(含代码示例)

    什么是内部类 什么是内部类? 顾名思义,就是将一个类的定义放在另一个类的内部. 概念很清楚,感觉很简单,其实关键在于这个内部类放置的位置,可以是一个类的作用域范围.一个方法的或是一个代码块的作用域范围 ...

  4. java-web调用后台下载方法

    后台下载指定文件必定会用到流, 无论使用poi还是使用jxl导出excel都需要用到流一种是outputstrean,另一种fileoutputstream第一种:如果想要弹出保存的提示框必须加入下列 ...

  5. dubbokeeper-moniter部署指南

    moniter在整个dubbo架构中的角色: 使用的1.0.1版本: ## 1.0.1版本变动内容 dubbokeeper在1.0.1版本对监控数据存储模块抽离出来,做为单独的应用部署,而不是和1.0 ...

  6. ZooKeeper系列(二)—— Zookeeper 单机环境和集群环境搭建

    一.单机环境搭建 1.1 下载 下载对应版本 Zookeeper,这里我下载的版本 3.4.14.官方下载地址:https://archive.apache.org/dist/zookeeper/ # ...

  7. Spring系列(四):Spring AOP详解

    一.AOP是什么 AOP(面向切面编程),可以说是一种编程思想,其中的Spring AOP和AspectJ都是现实了这种编程思想.相对OOP(面向过程编程)来说,提供了另外一种编程方式,对于OOP过程 ...

  8. python小白手册之远程链接转换

    访问顺序补充

  9. int string类型互转

    int -> String int i=12345;String s="";第一种方法:s=i+"";第二种方法:s=String.valueOf(i); ...

  10. 100天搞定机器学习|day44 k均值聚类数学推导与python实现

    [如何正确使用「K均值聚类」? 1.k均值聚类模型 给定样本,每个样本都是m为特征向量,模型目标是将n个样本分到k个不停的类或簇中,每个样本到其所属类的中心的距离最小,每个样本只能属于一个类.用C表示 ...