codeforce 933A Twisty Movement(dp+LIS)

A. A Twisty Movementtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputA 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.InputThe 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).OutputPrint a single integer, which means the maximum possible length of the longest non-decreasing subsequence of the new sequence.ExamplesinputCopy

4
1 2 1 2

output

4

inputCopy

10
1 1 2 2 2 1 1 2 2 1

output

9

NoteIn 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]
-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]
-dp1[i][j]+dp2[i][j]);
cout<<maxx<<endl;
return 0;
}