Codeforces 934C - A Twisty Movement

最新推荐文章于 2020-09-12 20:43:10 发布

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最新推荐文章于 2020-09-12 20:43:10 发布

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分类专栏： dp cf

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本文探讨了一个经典的计算机科学问题——如何通过选择性地反转序列中的某个区间来最大化序列中非递减子序列的长度。文章提供了一种有效的动态规划解决方案，并附带了完整的代码实现。

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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 
        
4
1 2 1 2

          output 
        
4

          input 
        
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：求最长不递减子序列可以选择一个区间逆转。

题解：求出1的前缀和，2的后缀和，以及区间[i,j]的最长不递增子序列。

dp[i][j][0]表示区间i-j以1结尾的最长不递增子序列；

dp[i][j][1]表示区间i-j以2结尾的最长不递增子序列，显然是区间i-j2的个数；

所以转移方程为：dp[i][j][1] = dp[i][j-1][1] + (a[j]==2);
dp[i][j][0] = max(dp[i][j-1][0], dp[i][j-1][1]) + (a[j]==1);（1<=i<=n,i<=j<=n）

#include <bits/stdc++.h>

using namespace std;
#define N 3000
#define LL long long
#define INF 0x3f3f3f3f
//const LL INF = 1e18;

LL a[N], b[N];
int dp1[N], dp2[N];
int sum1[N], sum2[N];
int dp[N][N][2];
int main()
{
    int n, t;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++)  {
        scanf("%d", &a[i]);
        sum1[i] += sum1[i-1] + (a[i]==1);
    }
    for(int i = n; i >= 1; i--)
        sum2[i] += sum2[i+1] + (a[i]==2);
    int ans = -INF;
    for(int i = 1; i <= n; i++) {
        int ans1=0, ans2=0;
        for(int j = i; j <= n; j++) {
            dp[i][j][1] = dp[i][j-1][1] + (a[j]==2);
            dp[i][j][0] = max(dp[i][j-1][0], dp[i][j-1][1]) + (a[j]==1);
            ans = max(ans, dp[i][j][1] + sum1[i-1]+sum2[j+1]);
            ans = max(ans, dp[i][j][0] + sum1[i-1]+sum2[j+1]);
        }
    }
    printf("%d\n", ans);
    return 0;
}