uva 10534 Wavio Sequence nlogn方法

Wavio Sequence
Input:StandardInput

Output: Standard Output

Time Limit: 2 Seconds

 

Wavio is a sequence of integers. It has some interestingproperties.

·  Wavio is of odd length i.e. L = 2*n+ 1.

·  The first (n+1) integers ofWavio sequence makes a strictly increasing sequence.

·  The last (n+1) integers of Waviosequence makes a strictly decreasing sequence.

·  No two adjacent integers are same in aWavio sequence.

For example 1, 2, 3, 4, 5, 4, 3, 2, 0 is an Waviosequence of length 9. But 1, 2, 3, 4, 5, 4, 3, 2, 2 is not avalid wavio sequence. In this problem, you will be given a sequence ofintegers. You have to find out the length of the longest Wavio sequence whichis a subsequence of the given sequence. Consider, the given sequence as :

1 2 3 2 1 2 3 4 32 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 outputwill be 9.

 

Input

The input filecontains less than 75 test cases. The description of each test case isgiven below: Input is terminated by end of file.

 

Each set starts with a postiveinteger, N(1<=N<=10000). In next few lines there will be Nintegers.

 

Output

For each set of input print the length of longest waviosequence in a line.

Sample Input                                  Output for 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

9 9 1

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

计算每个点的最长增序列和最长减序列

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include <queue>

using namespace std;
#define maxn 10005
#define inf 0x7ffffff
pair <int ,int > p;
int arr[maxn];
int LIS[maxn];
int LDS[maxn];
int dp[maxn];
int n;
//int cmpD(Node a,Node b){
//    return a.h > b.h;
//}
//int cmpI(Node a,Node b){
//    return a.h < b.h;
//}
int main()
{
    while(scanf("%d",&n) != EOF){
        for(int i = 1; i <= n; i++){
            scanf("%d",&arr[i]);
        }
        fill(dp,dp+n+1,inf);
        dp[0] = -inf +1;
        for(int i = 1; i <= n; i++){
            LIS[i] = lower_bound(dp,dp+n,arr[i])-dp;
            dp[LIS[i]] = arr[i];
        }
        fill(dp,dp+n+1,inf);
        dp[0] = -inf +1;
        for(int i = n; i >= 0; i--){
            LDS[i] = lower_bound(dp,dp+n,arr[i])-dp;
            dp[LDS[i]] = arr[i];
        }
        int ans = -inf;
        for(int i = 1; i <= n; i++){
            ans = max(ans, min(LIS[i],LDS[i]));
        }
        printf("%d\n",2*ans - 1);
    }
    return 0;
}