codeforce 933ATwisty Movement(dp+LIS)

A. A Twisty Movement time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

A dragon symbolizes wisdom, power and wealth. On Lunar New Year's Day, people model a dragon with bamboo strips and clothes, raise them with rods, and hold the rods high and low to resemble a flying dragon.

A performer holding the rod low is represented by a 1, while one holding it high is represented by a 2. Thus, the line of performers can be represented by a sequence a1, a2, ..., an.

Little Tommy is among them. He would like to choose an interval [l, r] (1 ≤ l ≤ r ≤ n), then reverse al, al + 1, ..., ar so that the length of the longest non-decreasing subsequence of the new sequence is maximum.

A non-decreasing subsequence is a sequence of indices p1, p2, ..., pk, such that p1 < p2 < ... < pk and ap1 ≤ ap2 ≤ ... ≤ apk. The length of the subsequence is k.

Input

The first line contains an integer n (1 ≤ n ≤ 2000), denoting the length of the original sequence.

The second line contains n space-separated integers, describing the original sequence a1, a2, ..., an (1 ≤ ai ≤ 2, i = 1, 2, ..., n).

Output

Print a single integer, which means the maximum possible length of the longest non-decreasing subsequence of the new sequence.

Examples input Copy 4 1 2 1 2 output 4 input Copy 10 1 1 2 2 2 1 1 2 2 1 output 9 Note

In the first example, after reversing [2, 3], the array will become [1, 1, 2, 2], where the length of the longest non-decreasing subsequence is 4.

In the second example, after reversing [3, 7], the array will become [1, 1, 1, 1, 2, 2, 2, 2, 2, 1], where the length of the longest non-decreasing subsequence is 9.

题意：给出n个数（均为1或2），求反转一个区间后，最大可以得到最长非减子序列的长度

要求最长非减子序列，首先想到LIS,可以正向和反向分别预处理出每一个区间的最长非减子序列长度，设正向为dp1[i][j],反向为dp2[i][j],那么所求即为max(dp1[1][n]-dp1[i][j]+dp2[i][j]),枚举i和j即可

#include<bits/stdc++.h> using namespace std; int a[2005]; int dp1[2005][2005],dp2[2005][2005],tmp[2005]; int main() { int n,len,maxx=0; scanf("%d",&n); for(int i=1; i<=n; i++) scanf("%d",&a[i]); for(int i=1; i<=n; i++) { len=0; memset(tmp,0,sizeof(tmp)); for(int j=i; j<=n; j++) { if(a[j]>=tmp[len]) tmp[++len]=a[j]; else { int pos=upper_bound(tmp+1,tmp+len+1,a[j])-tmp; tmp[pos]=a[j]; } dp1[i][j]=len; } } for(int i=n; i>=1; i--) { len=0; memset(tmp,0,sizeof(tmp)); for(int j=i; j>=1; j--) { if(a[j]>=tmp[len]) tmp[++len]=a[j]; else { int pos=upper_bound(tmp+1,tmp+len+1,a[j])-tmp; tmp[pos]=a[j]; } dp2[j][i]=len; } } for(int i=1;i<=n;i++) for(int j=i;j<=n;j++) maxx=max(maxx,dp1[1][n]-dp1[i][j]+dp2[i][j]); cout<<maxx<<endl; return 0; }