hdu 4604 Deque-优快云博客

本文链接：https://blog.youkuaiyun.com/zz_1215/article/details/38091591

本文详细介绍了在解决最长上升子序列与最长递减子序列问题时的一种高效方法，避免了重复计算，并通过实例展示了如何在实际应用中运用这种改进算法。

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最长上升子序列+最长递减子序列-重复的方法不严谨，貌似有人已经找到反例了，至于为什么那种方法能ac应该是测试数据弱吧

以下才是最标准的做法

//#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<cstring>
#include<iomanip>
#include<ctime>
using namespace std;
#ifdef _WIN32
typedef __int64 i64;
#define out64 "%I64d\n"
#define in64 "%I64d"
#else
typedef long long i64;
#define out64 "%lld\n"
#define in64 "%lld"
#endif
/************ for topcoder by zz1215 *******************/
#define foreach(c,itr)  for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
#define FOR(i,a,b)      for( int i = (a) ; i <= (b) ; i ++)
#define FF(i,a)         for( int i = 0 ; i < (a) ; i ++)
#define FFD(i,a,b)      for( int i = (a) ; i >= (b) ; i --)
#define S64(a)          scanf(in64,&a)
#define SS(a)           scanf("%d",&a)
#define LL(a)           ((a)<<1)
#define RR(a)           (((a)<<1)+1)
#define pb              push_back
#define pf              push_front
#define X               first
#define Y               second
#define CL(Q)           while(!Q.empty())Q.pop()
#define MM(name,what)   memset(name,what,sizeof(name))
#define MC(a,b)		memcpy(a,b,sizeof(b))
#define MAX(a,b)        ((a)>(b)?(a):(b))
#define MIN(a,b)        ((a)<(b)?(a):(b))
#define read            freopen("in.txt","r",stdin)
#define write           freopen("out.txt","w",stdout)

const int inf = 0x3f3f3f3f;
const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double oo = 10e9;
const double eps = 10e-9;
const double pi = acos(-1.0);
const int maxn = 101111;

int n;
int a[maxn];
vector<int>v;

inline void gao(int now)
{
	if (v.empty() || now >= v[v.size() - 1]){
		v.push_back(now);
	}
	else {
		v[upper_bound(v.begin(), v.end(), now) - v.begin()] = now;
	}
}

int main()
{
	int T;
	cin >> T;

	while (T--){
		cin >> n;
		for (int i = 1; i <= n; i++) {
			//cin >> a[i];
			SS(a[i]);
		}
		v.clear();
		int now;
		for (int i = n; i >= 1; i--) {
			now = 2 * a[i] + 1;
			gao(now);
		}
		for (int i = 1; i <= n; i++) {
			now = 2 * a[i];
			gao(now);
		}
		cout << v.size() << endl;
	}
	return 0;
}