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A sequence is provided. Please find two subsequences that are not handed over and are complete, so that the absolute sum of the Number Difference between two adjacent sub-sequences is the most

A sequence is provided. Please find two subsequences that are not handed over and are complete, so that the absolute sum of the Number Difference between two adjacent sub-sequences is the most

Question:

Given a sequence, please find two subsequences that are not handed over and are complete sets, so that the sum of the absolute values of the Number Difference between two adjacent sub-sequences is the smallest.

Ideas:

DP ..

Dp [I] [j] indicates that the last position of a person is I, and the last position of a person is j.

There are two situations:

1. i-j> 1: dp [I] [j] = dp [I-1] [j] + abs (pitch [I-2]-pitch [I-1]) keep A number ..

2. I-j = 1: 1) A takes only I, and all others are given to B.

2) A takes j and enumerates the position k obtained last time by A, B Takes I, and the last time k: res = min (res, dp [j] [k] + abs (I-1]-pitch [k-1])

Understanding: two people will regard one of the positions as the last position, and the last position of the other person is the last number. you can think of three people first. In this way, all the situations can be covered, which is equivalent to the subproblem of the problem.

AC.

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                        using namespace std; #define PB push_back #define MP make_pair #define REP(i,n) for(i=0;i<(n);++i) #define FOR(i,l,h) for(i=(l);i<=(h);++i) #define FORD(i,h,l) for(i=(h);i>=(l);--i) typedef vector
                       
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                            PII; int dp[2005][2005]; class SingingEasy { public: int solve(vector 
                           
                             pitch) { int len = pitch.size(); if(len <= 2) return 0; dp[1][0] = 0; for(int j = 2; j <= len; ++j) { dp[j][0] = dp[j-1][0] + abs(pitch[j-2]-pitch[j-1]); } dp[2][1] = 0; for(int i = 3; i <= len; ++i) { for(int j = 1; j < i; ++j) { if(i - j > 1) { dp[i][j] = dp[i-1][j] + abs(pitch[i-2]-pitch[i-1]); } else { int res = 1e9+5; for(int k = 1; k < j; ++k) { res = min(res, dp[j][k] + abs(pitch[i-1]-pitch[k-1])); } dp[i][j] = min(res, dp[j][0]); } } } int ans = dp[len][0]; for(int i = 1; i < len; ++i) { ans = min(ans, dp[len][i]); } return ans; }