Codeforces 934 C.A Twisty Movement-前缀和+后缀和+动态规划

C. 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.

 

这个题就是找的是反转区间之后的最长非递增子序列，并不是连续的。。。

其实就是把区间分成三部分，前一部分统计1的个数，后一部分统计2的个数，中间那一部分反转然后找最长的就可以。

前缀和记录1的个数，后缀和记录2的个数，动态规划操作中间那一部分找最长的。

 

代码:

 1 //C-用dp写
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<cmath>
 6 #include<algorithm>
 7 #include<cstdlib>
 8 using namespace std;
 9 const int maxn=2000+10;
10 int dp[maxn][maxn][2];
11 int p1[maxn],p2[maxn],a[maxn];
12 int main(){
13     int n;
14     scanf("%d",&n);
15     p1[0]=0;p2[n+1]=0;
16     for(int i=1;i<=n;i++){
17         scanf("%d",&a[i]);
18         p1[i]=p1[i-1];
19         if(a[i]==1)p1[i]++;
20     }
21     for(int j=n;j>=1;j--){
22         p2[j]=p2[j+1];
23         if(a[j]==2)p2[j]++;
24     }
25     int ans=-1;
26     for(int i=1;i<=n;i++){
27         for(int j=i;j<=n;j++){
28             if(a[j]==2)dp[i][j][1]=dp[i][j-1][1]+1;
29             else dp[i][j][1]=dp[i][j-1][1];
30             if(a[j]==1)dp[i][j][0]=max(dp[i][j-1][0],dp[i][j-1][1])+1;
31             else dp[i][j][0]=max(dp[i][j-1][0],dp[i][j-1][1]);
32             ans=max(p1[i-1]+p2[j+1]+dp[i][j][1],max(ans,p1[i-1]+p2[j+1]+dp[i][j][0]));
33         }
34     }
35     printf("%d\n",ans);
36 }

不用动态规划也可以直接模拟。

继续。