E. LIS of Sequence

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/486/problem/E

Description

The next "Data Structures and Algorithms" lesson will be about Longest Increasing Subsequence (LIS for short) of a sequence. For better understanding, Nam decided to learn it a few days before the lesson.

Nam created a sequence a consisting of n (1 ≤ n ≤ 105) elements a1, a2, ..., an (1 ≤ ai ≤ 105). A subsequence ai1, ai2, ..., aik where 1 ≤ i1 < i2 < ... < ik ≤ n is called increasing if ai1 < ai2 < ai3 < ... < aik. An increasing subsequence is called longest if it has maximum length among all increasing subsequences.

Nam realizes that a sequence may have several longest increasing subsequences. Hence, he divides all indexes i (1 ≤ i ≤ n), into three groups:

group of all i such that ai belongs to no longest increasing subsequences.
    group of all i such that ai belongs to at least one but not every longest increasing subsequence.
    group of all i such that ai belongs to every longest increasing subsequence.

Since the number of longest increasing subsequences of a may be very large, categorizing process is very difficult. Your task is to help him finish this job.

Input

The first line contains the single integer n (1 ≤ n ≤ 105) denoting the number of elements of sequence a.

The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 105).

Output

Print a string consisting of n characters. i-th character should be '1', '2' or '3' depending on which group among listed above index i belongs to.

Sample Input

4
1 3 2 5

Sample Output

3223

HINT

题意

给你n个数

然后问你这里面的每个数,是否是

1.不属于任何最长上升子序列中

2.属于多个最长上升子序列中

3.唯一属于一个最长上升子序列中

题解:

对于每一个数,维护两个dp

dp1表示1到i的最长上升子序列长度

dp2表示从n到i最长递减子序列长度

然后如果dp1[i]+dp2[i] - 1 == lis ,就说明属于lis里面,如果dp1[i]的值是唯一的,就说明唯一属于一个lis

否则就不属于咯

代码

#include<iostream>
#include<stdio.h>
#include<map>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 100005
int b[maxn];
int a[maxn];
void add(int x,int val)
{
while(x<=)
{
b[x] = max(b[x],val);
x += x & (-x);
}
}
int get(int x)
{
int ans = ;
while(x)
{
ans = max(ans,b[x]);
x -= x & (-x);
}
return ans;
}
int dp1[maxn];
int dp2[maxn];
int ans[maxn];
map<int,int> H;
int main()
{
int n;scanf("%d",&n);
int LIS = ;
for(int i=;i<=n;i++)
{
scanf("%d",&a[i]);
dp1[i] = + get(a[i]-);
add(a[i],dp1[i]);
LIS = max(LIS,dp1[i]);
}
reverse(a+,a++n);
memset(b,,sizeof(b));
for(int i=;i<=n;i++)
{
a[i] = - a[i] + ;
dp2[i] = + get(a[i] - );
add(a[i],dp2[i]);
}
reverse(dp2+,dp2++n);
for(int i=;i<=n;i++)
{
if(dp1[i]+dp2[i]-!=LIS)ans[i]=;
else H[dp1[i]]++;
}
for(int i=;i<=n;i++)
{
if(ans[i]!=&&H[dp1[i]]==)
{
ans[i]=;
}
}
for(int i=;i<=n;i++)
if(ans[i]==)
cout<<"";
else if(ans[i]==)
cout<<"";
else if(ans[i]==)
cout<<"";
}
/*
10
2 2 2 17 8 9 10 17 10 5
*/

Codeforces Round #277 (Div. 2) E. LIS of Sequence DP的更多相关文章

  1. Codeforces Round #277 (Div. 2) 题解

    Codeforces Round #277 (Div. 2) A. Calculating Function time limit per test 1 second memory limit per ...

  2. 【codeforces】Codeforces Round #277 (Div. 2) 解读

    门户:Codeforces Round #277 (Div. 2) 486A. Calculating Function 裸公式= = #include <cstdio> #include ...

  3. 贪心+构造 Codeforces Round #277 (Div. 2) C. Palindrome Transformation

    题目传送门 /* 贪心+构造:因为是对称的,可以全都左一半考虑,过程很简单,但是能想到就很难了 */ /************************************************ ...

  4. Codeforces Round #367 (Div. 2) C. Hard problem(DP)

    Hard problem 题目链接: http://codeforces.com/contest/706/problem/C Description Vasiliy is fond of solvin ...

  5. 套题 Codeforces Round #277 (Div. 2)

    A. Calculating Function 水题,分奇数偶数处理一下就好了 #include<stdio.h> #include<iostream> using names ...

  6. Codeforces Round #277(Div 2) A、B、C、D、E题解

    转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud A. Calculating Function 水题,判个奇偶即可 #includ ...

  7. Codeforces Round #277 (Div. 2)

    整理上次写的题目: A: For a positive integer n let's define a function f: f(n) =  - 1 + 2 - 3 + .. + ( - 1)nn ...

  8. Codeforces Round #277 (Div. 2) D. Valid Sets 暴力

    D. Valid Sets Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/486/problem ...

  9. Codeforces Round #277 (Div. 2) B. OR in Matrix 贪心

    B. OR in Matrix Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/486/probl ...

随机推荐

  1. 【转】IOS 计时器 NSTimer

    原文网址:http://blog.csdn.net/tangshoulin/article/details/7644124 1.初始化 + (NSTimer *)timerWithTimeInterv ...

  2. Ruby网络服务

    #require 'net/http' #p Net::HTTP.get_response(URI.parse('http://www.kuaiyoujia.com')){|res| # puts r ...

  3. VS2013密匙

    在网上找到的,亲测有用: BWG7X-J98B3-W34RT-33B3R-JVYW9

  4. iOS - 操作文件目录的方法

    转:http://blog.csdn.net/marujunyy/article/details/11579183 使用目录的常用方法: //获取当前目录 - (NSString *)currentD ...

  5. ch03-文字版面的设计

    Ch03: 文字版面的编辑 3.1 版面控制标记 3.1.1 取消文字换行: <NOBR> 1-取消换行文字实例:1-取消换行文字实例; 1-取消换行文字实例; 2-取消换行文字实例:2- ...

  6. oracle返回多结果集

    kavy 原文 oracle返回多结果集 Oracle存储过程: create or replace procedure P_Sel_TopCount2(in_top in number, out_c ...

  7. EasyHook远注简单监控示例 z

    http://www.csdn 123.com/html/itweb/20130827/83559_83558_83544.htm 免费开源库EasyHook(inline hook),下面是下载地址 ...

  8. Hie with the Pie(POJ 3311状压dp)

    题意:披萨店给n个地方送披萨,已知各地方(包括披萨店)之间花费的时间,求送完所有地方并回到店花费的最小时间 分析:状态好确定dp[i][j],i中1表示地方已送过,否则为0,j为当前状态最后一个送过的 ...

  9. HDU 4609 3-idiots FFT+容斥

    一点吐槽:我看网上很多分析,都是在分析这个题的时候,讲了半天的FFT,其实我感觉更多的把FFT当工具用就好了 分析:这个题如果数据小,统计两个相加为 x 的个数这一步骤(这个步骤其实就是求卷积啊),完 ...

  10. IOS PUSH 实践操作~~~~

    1.推送过程简介        (1)App启动过程中,使用UIApplication::registerForRemoteNotificationTypes函数与苹果的APNS服务器通信,发出注册远 ...