BadNeighbors

Problem Statement

A sequence of numbers is called a zig-zag sequence if the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with fewer than two elements is trivially a zig-zag sequence.

For example, 1,7,4,9,2,5 is a zig-zag sequence because the differences (6,-3,5,-7,3) are alternately positive and negative. In contrast, 1,4,7,2,5 and 1,7,4,5,5 are not zig-zag sequences, the first because its first two differences are positive and the second because its last difference is zero.

Given a sequence of integers, sequence, return the length of the longest subsequence of sequence that is a zig-zag sequence. A subsequence is obtained by deleting some number of elements (possibly zero) from the original sequence, leaving the remaining elements in their original order.

Definition

Class: ZigZag
Method: longestZigZag
Parameters: int[]
Returns: int
Method signature: int longestZigZag(int[] sequence)
(be sure your method is public)

Constraints
- sequence contains between 1 and 50 elements, inclusive.
- Each element of sequence is between 1 and 1000, inclusive.

Examples
0)

{ 1, 7, 4, 9, 2, 5 }
Returns: 6
The entire sequence is a zig-zag sequence.
1)

{ 1, 17, 5, 10, 13, 15, 10, 5, 16, 8 }
Returns: 7
There are several subsequences that achieve this length. One is 1,17,10,13,10,16,8.
2)

{ 44 }
Returns: 1
3)

{ 1, 2, 3, 4, 5, 6, 7, 8, 9 }
Returns: 2
4)

{ 70, 55, 13, 2, 99, 2, 80, 80, 80, 80, 100, 19, 7, 5, 5, 5, 1000, 32, 32 }
Returns: 8
5)

{ 374, 40, 854, 203, 203, 156, 362, 279, 812, 955,
600, 947, 978, 46, 100, 953, 670, 862, 568, 188,
67, 669, 810, 704, 52, 861, 49, 640, 370, 908,
477, 245, 413, 109, 659, 401, 483, 308, 609, 120,
249, 22, 176, 279, 23, 22, 617, 462, 459, 244 }
Returns: 36

题意：从a[0]-a[n-1]中取子序列，取得数不能相邻，a[0]和a[n-1]也算相邻。求这样的子序列和最大为多少。
状态dp[i]（前i个数中以a[i]结尾的最大不相邻子序列和）
由于a[0]和a[n-1]也算相邻，所以我把状态改成dp[j][i]（j={0,1},等于0时表示能取a[0]不取a[n-1]，等于1则反过来）

int ans(int*a,int n){    for(int i=0;i<n;i++){        if(i>0)dp[1][i]=a[i];        if(i<n-1)dp[0][i]=a[i];        for(int j=0;j+1<i;j++){                if(i>0)dp[1][i]=max(dp[1][i],dp[1][j]+a[i]);                if(i<n-1)dp[0][i]=max(dp[0][i],dp[0][j]+a[i]);        }    }    int ans=0;    for(int i=0;i<2;i++)        for(int j=0;j<n;j++)            ans=max(ans,dp[i][j]);    return ans;}