标签：des style blog http io ar color os sp
这题是要找一个最长（假设长度为2N-1）的子序列,使得前N个元素递增，后N个元素递减。
由于N比较大，直接上n^2的dp会超时……
用另外的方法……贪心+二分……这应该不算dp了……
好吧……也许可用斜率dp解？额……我不会
最长上升子序列问题：
给出一个由n个数组成的序列x[1..n]，找出它的最长单调上升子序列。即求最大的m和a1,
a2……,am，使得a1<a2<……<am且x[a1]<x[a2]<……<x[am]。
动态规划求解思路分析：（O(n^2)）
经典的O(n^2)的动态规划算法，设A[i]表示序列中的第i个数，F[i]表示从1到i这一段中以i结尾的最长上升子序列的长度，初始时设F[i] = 0(i = 1, 2, ..., len(A))。则有动态规划方程：F[i] = max{1, F[j] + 1} (j = 1, 2, ..., i - 1, 且A[j] < A[i])。
贪心+二分查找：（O(nlogn)）
开辟一个栈，每次取栈顶元素s和读到的元素a做比较，如果a>s， 则加入栈；如果a<s，则二分查找栈中的比a大的第1个数，并替换。 最后序列长度为栈的长度。
这也是很好理解的，对x和y，如果x<y且E[y]<E[x],用E[x]替换 E[y],此时的最长序列长度没有改变但序列Q的‘‘潜力‘‘增大。
举例：原序列为1，5，8，3，6，7
栈为1，5，8，此时读到3，则用3替换5，得到栈中元素为1，3，8， 再读6，用6替换8，得到1，3，6，再读7，得到最终栈为1，3，6，7 ，最长递增子序列为长度4。
#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
int a[10010];
int dp1[10010];
int dp2[10010];
int main()
{
int i,n,t;
vector<int>box;
while(cin>>n)
{
for(i=0;i<n;i++)
cin>>a[i];
box.clear();
for(i=0;i<n;i++)
{
if(box.empty()||a[i]>box.back())
{
box.push_back(a[i]);
dp1[i]=box.size();
}
else
{
t=lower_bound(box.begin(),box.end(),a[i])-box.begin();
box[t]=a[i];
dp1[i]=t+1;
}
}
box.clear();
for(i=n-1;i>-1;i--)
{
if(box.empty()||a[i]>box.back())
{
box.push_back(a[i]);
dp2[i]=box.size();
}
else
{
t=lower_bound(box.begin(),box.end(),a[i])-box.begin();
box[t]=a[i];
dp2[i]=t+1;
}
}
t=0;
for(i=0;i<n;i++)
t=max(t,min(dp1[i],dp2[i]));
cout<<t*2-1<<endl;
}
return 0;
}
| Time Limit: 3000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
Description
Problem D
Wavio Sequence
Input: Standard Input
Output: Standard Output
Time Limit: 2 Seconds
Wavio is a sequence of integers. It has some interesting properties.
· Wavio is of odd length i.e. L = 2*n + 1.
· The first (n+1) integers of Wavio sequence makes a strictly increasing sequence.
· The last (n+1) integers of Wavio sequence makes a strictly decreasing sequence.
· No two adjacent integers are same in a Wavio sequence.
For example 1, 2, 3, 4, 5, 4, 3, 2, 0 is an Wavio sequence of length 9. But 1, 2, 3, 4, 5, 4, 3, 2, 2 is not a valid wavio sequence. In this problem, you will be given a sequence of integers. You have to find out the length of the longest Wavio sequence which is a subsequence of the given sequence. Consider, the given sequence as :
1 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 1.
Here the longest Wavio sequence is : 1 2 3 4 5 4 3 2 1. So, the output will be
9.
Input
The input file contains less than 75 test cases. The description of each test case is given below: Input is terminated by end of file.
Each set starts with a postive integer, N(1<=N<=10000). In next few lines there will be N integers.
Output
For each set of input print the length of longest wavio sequence in a line.
10 1 2 3 4 5 4 3 2 1 10 19 1 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 1 5 1 2 3 4 5 |
9 9 1
|
Problemsetter: Md. Kamruzzaman, Member of Elite Problemsetters‘ Panel
Source
标签：des style blog http io ar color os sp
原文地址：http://blog.csdn.net/stl112514/article/details/41776399
