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. 基于zookeeper集群的云平台-配置中心的功能设计

    最近准备找工作面试,就研究了下基于zookeeper集群的配置中心. 下面是自己设想的关于开源的基于zookeeper集群的云平台-配置中心的功能设计.大家觉得哪里有问题,请提出宝贵的意见和建议,谢谢 ...

  2. javascript+jQuery补充

    一.jQuery事件绑定 <div class='c1'> <div> <div class='title'>菜单一</div> <div cla ...

  3. Apache之——多虚拟主机多站点配置的两种实现方案

    Apache中配置多主机多站点,可以通过两种方式实现: 将同一个域名的不同端口映射到不同的虚拟主机,不同端口映射到不同的站点: 将同一个端口映射成不同的域名,不同的域名映射到不同的站点. 我们只需要修 ...

  4. 9-1、大型项目的接口自动化实践记录----数据库结果、JSON对比

    上一篇写了如何从DB获取预期.实际结果,这一篇分别对不同情况说下怎么进行对比. PS:这部分在JSON对比中也适用. 1.结果只有一张表,只有一条数据 数据格式:因为返回的是dicts_list的格式 ...

  5. 伪分布式Spark + Hive on Spark搭建

    Spark大数据平台有使用一段时间了,但大部分都是用于实验而搭建起来用的,搭建过Spark完全分布式,也搭建过用于测试的伪分布式.现在是写一遍随笔,记录一下曾经搭建过的环境,免得以后自己忘记了.也给和 ...

  6. JVM总结(三)

    JVM总结(3)Class文件,类加载机制.编译过程 Java编译器先把Java代码编译为存储字节码的Class文件,再通过Class文件进行类加载. Class类文件的结构 Java编译器可以把Ja ...

  7. kpm字符串匹配算法

    首先是简单的朴素匹配算法 /* * 返回子串t在主串s的位置,若不存在则返回0 */ public static int index(String s, String t) { int i = 0;/ ...

  8. Mybatis学习笔记之---动态sql中标签的使用

    动态Sql语句中标签的使用 (一)常用标签 1.<if> if标签通常用于WHERE语句中,通过判断参数值来决定是否使用某个查询条件, 他也经常用于UPDATE语句中判断是否更新某一个字段 ...

  9. ABAP-复制采购订单行项目到新的行

    FUNCTION zmm_fm_copy2new. *"------------------------------------------------------------------- ...

  10. S3C2440 移植最新5.2linux内核

    基于 移植uboot后. 1. 移植linux内核 1.1 下载源码 打开 https://www.kernel.org/ 直接肝最新的 5.2.8 下载完后,在ubuntu里解压备用. 1.2 搭建 ...