Alignment

题目

样例

主要保持从前面往后面是递增的，从后面往前面是递增的即可，即至多有一个极值。

#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#define mem(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define _for(i,a,b) for( int i=(a); i<(b); ++i)
#define _rep(i,a,b) for( int i=(a); i<=(b); ++i)
typedef long long ll;
using namespace std;
const int N = 1100;
const int nmax = 50010;
const double esp = 1e-8;
const double PI = 3.1415926;
double a[N];
int lr[N], rl[N];
int main()
{
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%lf", &a[i]);
	}
	for (int i = 0; i < n; i++) {  
		lr[i] = 1;
		for (int j = 0; j < i; j++) {
			if (a[j] < a[i]) {
				lr[i] = max(lr[i], lr[j] + 1);
			}
		}
	}
	for (int i = n - 1; i >= 0; i--) {
		rl[i] = 1;
		for (int j = n - 1; j > i; j--) {
			if (a[i] > a[j])
				rl[i] = max(rl[j] + 1, rl[i]);
		}
	}
	int maxl = 0;
	for (int i = 0; i < n; i++) {  
		for (int j = i + 1; j < n; j++) {
			maxl = max(maxl, lr[i] + rl[j]);
		}
	}
	printf("%d\n", n - maxl);
	return 0;
}