HOJ 2985 Wavio Sequence(最长递增子序列以及其O(n*logn)算法)
来源:互联网 发布:淘宝上有卖少女脚皮的 编辑:程序博客网 时间:2024/06/05 01:54
Wavio Sequence
My Tags (Edit)
Source : UVA
Time limit : 1 sec Memory limit : 32 M
Submitted : 296, Accepted : 123
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 multiple 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.
Sample Input
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
Sample Output
9
9
1
解法是将数组正着和倒着分别求一下最长递增子序列,然后遍历i,如果i左边的数组和右边的数组的最长子序列相等,就是符合条件的。如果求最长递增子序列用O(N^2)会超时,所以必须用效率高的算法。。关于O(n*logn)算法请参考以下博客
http://blog.csdn.net/dacc123/article/details/50571844
贴上代码
#include <iostream>#include <string.h>#include <stdlib.h>#include <math.h>#include <algorithm>using namespace std;int a[10005];int b[10005];int c[10005];int d[10005];int dp[10005];int bp[10005];int n;int ans;int search(int num,int l,int r,int *dp){ int mid; while(l<=r) { mid=(l+r)/2; if(num>dp[mid]) l=mid+1; else r=mid-1; } return l;}int main(){ while(scanf("%d",&n)!=EOF) { ans=0; for(int i=1;i<=n;i++) { scanf("%d",&a[i]); b[n-i+1]=a[i]; } //memset(dp,0,sizeof(dp)); dp[1]=a[1]; c[1]=1; int len=1; for(int i=2;i<=n;i++) { if(a[i]>dp[len]) dp[++len]=a[i]; else { int pos=search(a[i],1,len,dp); dp[pos]=a[i]; } c[i]=len; } bp[1]=b[1]; d[1]=1; int len2=1; for(int i=2;i<=n;i++) { if(b[i]>bp[len2]) bp[++len2]=b[i]; else { int pos=search(b[i],1,len2,bp); bp[pos]=b[i]; } d[i]=len2; } for(int i=1;i<=n;i++) { if(c[i]==d[n-i+1]) ans=max(ans,c[i]*2-1); } printf("%d\n",ans); } return 0;}
- HOJ 2985 Wavio Sequence(最长递增子序列以及其O(n*logn)算法)
- 杭电1257(最长递增子序列O(N*N)+O(N*logN))
- 最长上升子序列 o(n*logn)算法
- 51nod_1134 最长递增子序列(O(n*logn))
- 最长递增子序列问题,O(N*logN)实现
- 算法导论15.4-6 求一个n个数的序列的最长单调递增子序列 O(n*logn)
- 最长不下降子序列的O(n*logn)算法
- DP之最长上升子序列O(n*logn)算法
- DP之最长上升子序列O(n*logn)算法
- UVa 10534 Wavio Sequence (最长递增子序列 DP 二分)
- UVa 10534 Wavio Sequence ( DP 二分 最长递增子序列 )
- UVA - 10534 Wavio Sequence 最长递增子序列的nlogn算法
- HDU 1025 Constructing Roads (最长上升子序列O(n*logn)算法)
- 最长XX子序列N*LOGN算法
- uva 10534 Wavio Sequence (最大递增和递减序列 n*logn)
- uva10534 - Wavio Sequence O(nlgn)的最长上升子序列
- UVA-10534-Wavio Sequence(最长单调递增子序列长度NlogN)
- nyoj17单调递增最长子序列(N*logN)
- 2015年第六届蓝桥杯C/C++程序设计本科B组省赛 牌型种数(结果填空)
- hdoj--5619--Jam's store(最小费用最大流)
- Redis与MySql结合
- 数据结构 栈 链表
- 百度2014笔试算法题
- HOJ 2985 Wavio Sequence(最长递增子序列以及其O(n*logn)算法)
- UILabel / UITextView load HTML文本
- Cocos2d-x 3.x数据存储(UserDefault)
- Leetcode_299_Bulls and Cows
- Android事件机制
- 通过java.net.URLConnection发送HTTP请求的方法
- hdu 4146 Flip Game
- Unity5 弹力球的制作
- 蓝桥杯 三羊献瑞