【Codeforces 1420 C1】Pokémon Army (easy version)，dp，奇偶性分类讨论

problem

C1. Pokémon Army (easy version)
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
This is the easy version of the problem. The difference between the versions is that the easy version has no swap operations. You can make hacks only if all versions of the problem are solved.

Pikachu is a cute and friendly pokémon living in the wild pikachu herd.

But it has become known recently that infamous team R wanted to steal all these pokémon! Pokémon trainer Andrew decided to help Pikachu to build a pokémon army to resist.

First, Andrew counted all the pokémon — there were exactly n pikachu. The strength of the i-th pokémon is equal to ai, and all these numbers are distinct.

As an army, Andrew can choose any non-empty subsequence of pokemons. In other words, Andrew chooses some array b from k indices such that 1≤b1<b2<⋯<bk≤n, and his army will consist of pokémons with forces ab1,ab2,…,abk.

The strength of the army is equal to the alternating sum of elements of the subsequence; that is, ab1−ab2+ab3−ab4+….

Andrew is experimenting with pokémon order. He performs q operations. In i-th operation Andrew swaps li-th and ri-th pokémon.

Note: q=0 in this version of the task.

Andrew wants to know the maximal stregth of the army he can achieve with the initial pokémon placement. He also needs to know the maximal strength after each operation.

Help Andrew and the pokémon, or team R will realize their tricky plan!

Input
Each test contains multiple test cases.

The first line contains one positive integer t (1≤t≤103) denoting the number of test cases. Description of the test cases follows.

The first line of each test case contains two integers n and q (1≤n≤3⋅105,q=0) denoting the number of pokémon and number of operations respectively.

The second line contains n distinct positive integers a1,a2,…,an (1≤ai≤n) denoting the strengths of the pokémon.

i-th of the last q lines contains two positive integers li and ri (1≤li≤ri≤n) denoting the indices of pokémon that were swapped in the i-th operation.

It is guaranteed that the sum of n over all test cases does not exceed 3⋅105, and the sum of q over all test cases does not exceed 3⋅105.

Output
For each test case, print q+1 integers: the maximal strength of army before the swaps and after each swap.

Example
inputCopy
3
3 0
1 3 2
2 0
1 2
7 0
1 2 5 4 3 6 7
outputCopy
3
2
9
Note
In third test case we can build an army in such way: [1 2 5 4 3 6 7], its strength will be 5−3+7=9.

solution

/*
题意：
+ 给出一个长为n（3e5）的序列。随机选k个数，最小化ab1-ab2+ab3-ab4...+abk的值
思路：
+ dp,维护奇数和偶数时的最大和，转移时对奇偶性分类讨论。
+ 即维护d1[i+1] = max(d1[i],d2[i]+a[i])，表示到i+1为止的最大和（i+1为奇数），d2则表示i+1位偶数。
*/
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 3e5+10;
int a[maxn];
int main(){
	ios::sync_with_stdio(false);
	int T;  cin>>T;
	while(T--){
		int n, q;  cin>>n>>q;
		for(int i = 1; i <= n; i++)
			cin>>a[i];
		for(int i = 1; i <= q; i++){
			int l, r;  cin>>l>>r;  swap(a[l],a[r]);
		}
		vector<LL>d1(n+2), d2(n+2);
		d1[1] = -1e9, d2[1] = 0;
		for(int i = 1; i <= n; i++){
			d1[i+1] = max(d1[i],d2[i]+a[i]);
			d2[i+1] = max(d2[i],d1[i]-a[i]);
		}
		cout<<max(d1.back(),d2.back())<<"\n";
	}
	return 0;
}
  
 

  
1
  
2
  
3
  
4
  
5
  
6
  
7
  
8
  
9
  
10
  
11
  
12
  
13
  
14
  
15
  
16
  
17
  
18
  
19
  
20
  
21
  
22
  
23
  
24
  
25
  
26
  
27
  
28
  
29
  
30
  
31
  
32
  
33
  
34
 

文章来源: gwj1314.blog.csdn.net，作者：小哈里，版权归原作者所有，如需转载，请联系作者。

原文链接：gwj1314.blog.csdn.net/article/details/113355164