PAT (Top Level) Practice 1017 The Best Peak Shape （35 分）

这道题的链接：https://pintia.cn/problem-sets/994805148990160896/problems/994805150479138816

这道题很明显是考最长递增序列，用LIS   O(NlgN）算法，从网上看了别人写的大多用dp但耗时太长，有用LIS的但却用的不够准确。在LIS时加一个数组nu 记录：到该下标的数时的最长递增序列为多少。

code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
using namespace std;
#define N 10005
int n;
int Bisearch(int len, int w,int *List) {
	int l = 1, r = len;
	int mid;
	while (l <= r) {
		mid = (l + r) / 2;
		if (List[mid] > w)
			r = mid - 1;
		else if (List[mid] < w)
			l = mid + 1;
		else
			return mid;
	}
	return l;
}
int LIS(int *a,int *List, int *nu ) {
	int len = 1;
	int pos;
	List[1] = a[1];
	nu[1] = 1;
	for (int i = 2; i <= n; i++) {
		if (a[i] > List[len]) {
			List[++len] = a[i];
			nu[i] = len;
		}
		else {
			pos = Bisearch(len, a[i], List);
			List[pos] = a[i];
			//cout <<i<<" "<< pos << endl;
			nu[i] = pos;
		}
	}
	return len;
}
int main() {
	int nu[N], nu1[N], nu2[N];
	int a[N], b[N], List[N], List1[N];
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
	}
	for (int i = n, j = 1; j <= n; j++, i--) {
		b[i] = a[j];
	}
	int lena = LIS(a, List, nu);
	if (lena == 1 || lena == n) {
		cout << "No peak shape" << endl;
		return 0;
	}
	int lenb = LIS(b, List1, nu1);
	for (int i = 1, j = n; i <= n; j--, i++)
	{
		nu2[i] = nu1[j];
	}
	int mx = -1;
	int cha = 10 * N;
	int t;
	for (int i = 1; i <= n; i++) {
		//cout << nu[i] << " " << nu2[i] << endl;
		if (nu[i] > 1 && nu2[i] > 1)
		{
			if (nu[i] + nu2[i] >= mx) {
				if (nu[i] + nu2[i] == mx) {
					if (abs(nu[i] - nu2[i]) <= cha) {
						t = i;
						cha = abs(nu[i] - nu2[i]);
					}
				}
				else {
					mx = nu[i] + nu2[i];
					t = i;
					cha = abs(nu[i] - nu2[i]);
				}
			}
		}
	}
	if (mx == -1) {
		cout << "No peak shape" << endl;
		return 0;
	}
	cout << mx - 1 << " " << t << " " << a[t] << endl;
	return 0;
}