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UVA 10534-Wavio Sequence(dp_正序逆序最長上升子序列)

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UVA 10534-Wavio Sequence(dp_正序逆序最長上升子序列)

Wavio Sequence Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu Submit Status

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.

Sample Input Output for Sample Input

1 2 3 4 5 4 3 2 1 10

1 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 1

1 2 3 4 5

Problemsetter: Md. Kamruzzaman, Member of Elite Problemsetters' Panel

Wavio是一個整數序列，具有以下特性：
1、Wavio序列的長度是奇數，即 L = 2 * n + 1;
2、Wavio序列前 n+1 個整數是遞增序列
3、Wavio序列後 n+1 個整數是遞減序列
如示例 1 2 3 4 5 4 3 2 1 10
最長的 Wavio序列為 1 2 3 4 5 4 3 2 1 ，所以答案為9
對於輸入序列中的一個整數 ai ，我們設以 ai 為尾的前綴的最長遞增序列的長度為Fi ，如在示例1中對已第3個整數3，從頭開始，以3為尾的遞增序列為1 2 3 ，所以F3=3；
以ai為首的後綴的遞減序列的長度為Gi, 如示例1中第3個整數，以3為開始，遞減序列為3 2 1，所以G3=3 (可以看做求從最後一個元素出發，到3這個位置的最大遞增序列，為1 2 3，所以G3=3)
在我們找遞減序列的時候，可以看做從最後一個元素出發，到當前位置的最大遞增序列，這樣我們對於每一個元素ai 我們可以先求出 a0到ai最大遞增序列和 an-1到ai的最大遞增序列，這樣我們可以得到ai的Wavio序列的值為
2*min( Fi , Gi) - 1 ,最後的結果是這些Wavio序列中的一個最大值。

#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;
const int inf=0x3f3f3f3f;
int binsearch(int *a,int r,int key)
{
    int l=0;
    int mid=(l+r)/2;
    while(l<=r){
        if(a[mid]key)
            r=mid-1;
        else
            return mid;
        mid=(l+r)/2;
    }
    return l;
}

int main()
{
    int n,i;
    while(~scanf("%d",&n)){
      int a[10010],b[10010],c[10010];//b保存的是以a[i]為結尾的最長遞增子序列，c保存長度為i的遞增子序列的最末的元素，c是單調遞增的
      int a1[10010],b1[10010],c1[10010];//同上
      for(i=0;i