1471 - Defense Lines
After the last war devastated your country, you - as the king of the land of Ardenia - decided it was
high time to improve the defense of your capital city. A part of your fortification is a line of mage
towers, starting near the city and continuing to the northern woods. Your advisors determined that the
quality of the defense depended only on one factor: the length of a longest contiguous tower sequence
of increasing heights. (They gave you a lengthy explanation, but the only thing you understood was
that it had something to do with firing energy bolts at enemy forces).
After some hard negotiations, it appeared that building new towers is out of question. Mages of
Ardenia have agreed to demolish some of their towers, though. You may demolish arbitrary number of
towers, but the mages enforced one condition: these towers have to be consecutive.
For example, if the heights of towers were, respectively, 5, 3, 4, 9, 2, 8, 6, 7, 1, then by demolishing
towers of heights 9, 2, and 8, the longest increasing sequence of consecutive towers is 3, 4, 6, 7.
Input
The input contains several test cases. The first line of the input contains a positive integer Z ≤ 25,
denoting the number of test cases. Then Z test cases follow, each conforming to the format described
below.
The input instance consists of two lines. The first one contains one positive integer n ≤ 2 · 105
denoting the number of towers. The second line contains n positive integers not larger than 109
separated by single spaces being the heights of the towers.
Output
For each test case, your program has to write an output conforming to the format described below.
You should output one line containing the length of a longest increasing sequence of consecutive
towers, achievable by demolishing some consecutive towers or no tower at all.
Sample Input
2
9
5 3 4 9 2 8 6 7 1
7
1 2 3 10 4 5 6
Output
4
6
解题思路:
本问题的关键在于set的动态更新,对set集合各种操作的熟练运用是关键。详细思路见紫书。
代码如下:
#include <iostream>
#include <set>
#include <map>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn=+;
int A[maxn];
int f[maxn];
int g[maxn];
int n;
map<int,int> m;
set<int> s; void pre_procession(){
int L=,R=;
while(L<n){
while(R<n&&(R==L||A[R]>A[R-])){g[R]=R+-L;R++;}
while(L<R){f[L]=R-L;L++;}
}
}
void putset(){
int ans=;
for(int i=;i<n;i++){
set<int>::iterator it=s.lower_bound(A[i]);
if(it!=s.begin()){
it--;
ans=max(ans,f[i]+m[*it]);
if(m[*it]>=g[i]) continue;
it++;
if(*it==A[i]&&m[*it]>=g[i]) continue;
}
s.insert(it, A[i]);
m[A[i]]=g[i];
int cur=A[i];
it=s.upper_bound(cur);
while(it!=s.end()&&m[cur]>=m[*it]){
cur=*it;
s.erase(it);
it=s.upper_bound(cur);
}
}
cout<<ans<<endl;
}
int main(int argc, const char * argv[]) {
int T;
scanf("%d",&T);
while(T--){
scanf("%d",&n);
for(int i=;i<n;i++) scanf("%d",&A[i]);
pre_procession();
m.clear();
s.clear();
putset();
}
return ;
}
1471 - Defense Lines的更多相关文章
- UVA - 1471 Defense Lines 树状数组/二分
Defense Lines After the last war devastated your country, you - as the ...
- UVa 1471 Defense Lines - 线段树 - 离散化
题意是说给一个序列,删掉其中一段连续的子序列(貌似可以为空),使得新的序列中最长的连续递增子序列最长. 网上似乎最多的做法是二分查找优化,然而不会,只会值域线段树和离散化... 先预处理出所有的点所能 ...
- uva 1471 Defense Lines
题意: 给一个长度为n(n <= 200000) 的序列,你删除一段连续的子序列,使得剩下的序列拼接起来,有一个最长的连续递增子序列 分析: 就是最长上升子序列的变形.需要加一个类似二分搜索就好 ...
- UVA - 1471 Defense Lines (set/bit/lis)
紫薯例题+1. 题意:给你一个长度为n(n<=200000)的序列a[n],求删除一个连续子序列后的可能的最长连续上升子序列的长度. 首先对序列进行分段,每一段连续的子序列的元素递增,设L[i] ...
- UVA 1471 Defense Lines 防线 (LIS变形)
给一个长度为n的序列,要求删除一个连续子序列,使剩下的序列有一个长度最大的连续递增子序列. 最简单的想法是枚举起点j和终点i,然后数一数,分别向前或向后能延伸的最长长度,记为g(i)和f(i).可以先 ...
- UVa 1471 Defense Lines (二分+set优化)
题意:给定一个序列,然后让你删除一段连续的序列,使得剩下的序列中连续递增子序列最长. 析:如果暴力枚举那么时间复杂度肯定受不了,我们可以先进行预处理,f[i] 表示以 i 结尾的连续最长序列,g[i] ...
- Uva 1471 Defense Lines(LIS变形)
题意: 给你一个数组,让你删除一个连续的子序列,使得剩下的序列中有最长上升子序列, 求出这个长度. 题解: 预处理:先求一个last[i],以a[i]为开始的合法最长上升子序列的长度.再求一个pre[ ...
- 【二分】Defense Lines
[UVa1471] Defense Lines 算法入门经典第8章8-8 (P242) 题目大意:将一个序列删去一个连续子序列,问最长的严格上升子序列 (N<=200000) 试题分析:算法1: ...
- UVa 1471 (LIS变形) Defense Lines
题意: 给出一个序列,删掉它的一个连续子序列(该子序列可以为空),使得剩下的序列有最长的连续严格递增子序列. 分析: 这个可以看作lrj的<训练指南>P62中讲到的LIS的O(nlogn) ...
随机推荐
- CSS3--关于z-index不生效问题
最近写CSS3和js结合,遇到了很多次z-index不生效的情况: 1.在用z-index的时候,该元素没有定位(static定位除外) 2.在有定位的情况下,该元素的z-index没有生效,是因为该 ...
- HTML-DOM常用对象的用法(select/option/form/table)
HTML DOM 常用对象: 它对常用HTML元素操作的简化. Select对象 它代表页面上的一个select元素,常用属性有: select.value ——当前选中项的value ,没有valu ...
- vue2-vux-fitness-project
非常感谢那些无私开源的程序员,希望我也能够有能力像你们那样,开源很多很有意思的东西~~ //index.html <!DOCTYPE html> <html> <head ...
- Effective Modern C++:02auto
05:优先使用auto,而非显示类型声明 显示类型声明有下面一些缺点: int x; //未初始化,或者初始化为0,视语境而定 template<typename It> void dwi ...
- LeetCode225 Implement Stack using Queues
Implement the following operations of a stack using queues. (Easy) push(x) -- Push element x onto st ...
- ANSI编码方式转化为UTF-8方式
说明: 记事本txt有四种编码方式,分别为:UTF-8.ANSI.Unicode和Unicode big endian,当进行写操作,创建的txt编码格式,与写入汉字的编码方式相同:如果写入的汉字是不 ...
- python-selenium自动化测试(火狐、谷歌、360浏览器启动)
一.打开谷歌浏览器 import selenium from selenium import webdriver browser = webdriver.Chrome(executable_path ...
- java代码简单实现队列
1. 基于链表简单实现 import lombok.AllArgsConstructor; import lombok.Data; import lombok.NoArgsConstructor; / ...
- C++讲课总结 标签: c++总结 2015-02-28 14:48 671人阅读 评论(25) 收藏
昨天老师算是给串了一本C++ 的课本,根据自己的理解,赶紧记录一下,也好作为自己学习时候的根据. C++编程简介:每本讲语言的书,第一章总是简介,内容无非是发展历史,语言特色等东西,专业的东西不多,都 ...
- etcd 在超大规模数据场景下的性能优化
作者 | 阿里云智能事业部高级开发工程师 陈星宇(宇慕) 概述 etcd是一个开源的分布式的kv存储系统, 最近刚被cncf列为沙箱孵化项目.etcd的应用场景很广,很多地方都用到了它,例如kuber ...