# Longest ascending subsequence nlogn

Source: Internet
Author: User

In the longest ascending subsequence, for the number of IPTS [I], traverse forward, when the number of IPTS [J] is smaller than the value of IPTS [I], IPTS [J] can be used as the first digit in the ascending sequence.

DP [I] = max {DP [J] + 1 | j <I & IPT [J] <IPT [I]}

If two statuses A and B meet the requirements of DP [a] = DP [B] and

It is better for the following status DP [a], because if you set the stage to "I"> "DP [B", you must set the stage to "IPT [I]>" DP [a] ". if you do not use the following command,

Therefore, if you only save status A, the optimal solution will not be lost.

For the same Dp value, you only need to keep one of the smallest

Then, the error message returned when the error message returned ).....

This sequence is ordered.

For example, 1 6 2 3 7 5

DP 1 2 2 3 4 4

When 2 is calculated, 6 can be replaced with 2 if the value of "2" is 6.

Similarly, 1 (6)2 3 (7)5

DP 1 2 3 4

In this way, you can maintain a sequence during computing.

`//#pragma comment(linker, "/STACK:102400000,102400000")//HEAD#include <cstdio>#include <cstring>#include <vector>#include <iostream>#include <algorithm>#include <queue>#include <string>#include <set>#include <stack>#include <map>#include <cmath>#include <cstdlib>using namespace std;//LOOP#define FE(i, a, b) for(int i = (a); i <= (b); ++i)#define FED(i, b, a) for(int i = (b); i>= (a); --i)#define REP(i, N) for(int i = 0; i < (N); ++i)#define CLR(A,value) memset(A,value,sizeof(A))//STL#define PB push_back//INPUT#define RI(n) scanf("%d", &n)#define RII(n, m) scanf("%d%d", &n, &m)#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)#define RS(s) scanf("%s", s)typedef long long LL;const int INF = 0x3f3f3f3f;const int MAXN = 1010;#define FF(i, a, b) for(int i = (a); i < (b); ++i)#define FD(i, b, a) for(int i = (b) - 1; i >= (a); --i)#define CPY(a, b) memcpy(a, b, sizeof(a))#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)#define EQ(a, b) (fabs((a) - (b)) <= 1e-10)#define ALL(c) (c).begin(), (c).end()#define SZ(V) (int)V.size()#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)#define WI(n) printf("%d\n", n)#define WS(s) printf("%s\n", s)#define sqr(x) (x) * (x)typedef vector <int> VI;typedef unsigned long long ULL;const double eps = 1e-10;const LL MOD = 1e9 + 7;int ipt, dp, a;int main(){    int n;    while (~RI(n))    {        REP(i, n)            RI(ipt[i]);        REP(i, n)   dp[i] = 1;        int ans = 0;        CLR(a, INF);        REP(i, n)        {            int x = lower_bound(a, a + ans, ipt[i]) - a;            dp[i] =  x + 1;            ans = max(ans, dp[i]);            a[x] = ipt[i];        }        cout << ans << endl;    }    return 0;}`

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