1421 - Wavio Sequence

题目大意:求一个序列中 先严格递增后严格递减的子序列的数目(要求这个子序列对称)。

题目思路:正一遍DP,反一遍DP,因为n<=1e5,dp要把时间压缩到nlogn

 

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define MAXSIZE 100005

using namespace std;

int dp[MAXSIZE],a[MAXSIZE];

int n;

int Solve()
{
    dp[1] = 1;
    int G1[MAXSIZE],G2[MAXSIZE],len=1,pos;
    G1[1] = a[1];
    for(int i=2;i<=n;i++)
    {
        if(a[i] > G1[len])
            pos = ++len;
        else
        {
            pos = lower_bound(G1+1,G1+1+len,a[i]) - (G1+1) + 1;
        }
        G1[pos] = a[i];
        dp[i] = len;
    }

    G2[1] = a[n];
    len = 1;
    int ans = 1;
    for(int i=n-1;i>=1;i--)
    {
        if(a[i] > G2[len])
            pos = ++len;
        else
            pos = lower_bound(G2+1,G2+1+len,a[i]) - (G2+1) + 1;
        G2[pos] = a[i];
        int temp = min(len,dp[i]);
        ans = max(ans,temp);
    }
    return 2*ans - 1;
}

int main()
{
    int T,cns=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        int ans =Solve();
        printf("Case %d: %d\n",cns++,ans);
    }
    return 0;
}
View Code

 

posted @ 2017-10-19 13:40  声声醉如兰  阅读(213)  评论(0编辑  收藏  举报