CF 1373D Maximum Sum on Even Positions （DP）

传送门

题目：给定长度为n的数组，下标从0开始。你能够至多翻转一次连续的子数组，问a0 + a2 + ... + a2k最大是多少。

思路：咱们发现一个状况： 20 30 10 ...，咱们发现若是30和20反转也能够和后面10的的反转，就分红了两种状况，咱们能够经过dp来解决，当前这个数与左边一块儿反转差值仍是与右边一块儿反转翻转差值大。

初始化dp[0~n] = 0，咱们把下标改为从1开始，则变成a1 + a3 + a5 + ... + a2k-1。

①i & 1 : dp[i] = max(dp[i], dp[i - 2] + a[i - 1] - a[i]

②！（i&1）: dp[i] = max(dp[i], dp[i - 2] + a[i] - a[i - 1)

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <queue>
 5 #include <string>
 6 #include <vector>
 7 #include <cmath>
 8  
 9 using namespace std;
10  
11 #define ll long long
12 #define pb push_back
13 #define fi first
14 #define se second
15  
16 const int N = 2e5 + 10;
17 ll a[N], dp[N], sum; 
18 
19 void solve()
20 {      
21     int T;
22     cin >> T;
23     while(T--){
24         int n;
25         cin >> n;
26 
27         sum = 0;
28         for(int i = 1; i <= n; ++i){
29             cin >> a[i];
30             if(i & 1) sum += a[i];
31         }
32         for(int i = 1; i <= n; ++i) dp[i] = 0;
33 
34         for(int i = 2; i <= n; ++i){
35             if(i & 1){
36                 dp[i] = max(dp[i], dp[i - 2] + a[i - 1] - a[i]);
37             }else{
38                 dp[i] = max(dp[i], dp[i - 2] + a[i] - a[i - 1]);
39             }
40         }
41 
42         ll max_d = 0;
43         for(int i = 0; i <= n; ++i) max_d = max(max_d, dp[i]);
44         //cout << "max = ";
45         cout << sum + max_d << endl;    
46     }
47 }
48  
49 int main()
50 {
51     ios::sync_with_stdio(false);
52     cin.tie(0);
53     cout.tie(0); 
54     solve();
55  
56     return 0;
57 }