"Simulation competition 20181025": Game Theory + dp simple Optimization

Description

Yasuo and riven start to practice for a row of \ (n \) dummy persons. Kill the \ (I \) dummy and get \ (C_ I. The two sides took turns and learned from each other in their exercises, so their swords became stronger and stronger. Based on the number of counterfeits killed by the peer \ (k \), you can kill the number of consecutive counterfeits in the range [1, 2 K] \ at the top of the remaining counterfeits. Yasuo first made a move to kill \ (1 \) or \ (2 \) counterfeits. Yasuo secretly called you aside and asked how much essence he could get when both parties adopt the best strategy.

Input

The first line is a positive integer \ (n \), indicating the number of dummy persons.
Next, for the \ (n \) Row, each row has a positive integer \ (C_ I \), which indicates the essence of killing each dummy.

Output

A positive integer indicates the maximum number of excellent values that Yasuo can obtain.

Sample Input

513172

Sample output

9

Example

Yasuo chopped \ (1 \), riven chopped \ (2 \), Yasuo chopped \ (3, 4 \), riven chopped \ (5.

Data range

For the previous \ (10 \ % \) data, \ (n \ Leq 10 \)
For the previous \ (40 \ % \) data, \ (n \ Leq 500 \)
For \ (100 \ % \) data, \ (5 \ Leq n \ Leq 5000, CI \ Leq 10 ^ 9 \)

Question

First, let's talk about the background of the question, and talk about the author who moved the question and made the magic change.

Simple game theory \ (DP \) involves almost no knowledge of game theory.
Obviously, the two are equivalent. \ (F [I] [J] \) indicates the \ (I \) dummy at the end of the current record, the last knife is the maximum value of \ (J \) dummy. Obviously, we can enumerate the number of people \ (k \ In [1, 2j] \) and \ (DP \) states to be transferred. Unfortunately, this complexity is \ (O (N ^ 3) \) and cannot pass all test points.
For optimization, let's write down the \ (DP \) format:
\ (F [I] [J] = max (s [I]-f [I-K] [k]) \), where \ (k \ In [1, 2j] \), \ (s [I] \) represents the sum of \ (I \) Personal \ (C.
The following formula is used:
\ (F [I] [J] = max (F [I] [J-1], s [I]-f [i-2j] [2j], s [I]-f [i-2j + 1] [2j-1]) \)
And then it's gone ......
\ (Code :\)

#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define N 5005#define ll long long#define inf (1ll << 50)template<typename Mytype>void Read(Mytype &p){    p = 0;    char c = getchar();    for (; c < '0' || c > '9'; c = getchar());    for (; c >= '0' && c <= '9'; c = getchar())p = p * 10 + c - '0';}ll s[N];ll f[5005][5005];int n, A[N];int main(){    Read(n);    for (int i = 1; i <= n; i++)        Read(A[i]), s[n - i + 1] = A[i];    for (int i = 1; i <= n; i++)        s[i] += s[i - 1];    for (int i = 1; i <= n; i++)    {        for (int j = 1; j <= n; j++)        {            ll ans1 = -inf, ans2 = -inf;            if (i >= (2 * j - 1))                ans1 = s[i] - f[i - (2 * j - 1)][2 * j - 1];            if (i >= (2 * j))                ans2 = s[i] - f[i - 2 * j][2 * j];            f[i][j] = max(max(ans1, ans2), f[i][j - 1]);        }    }    printf("%lld\n", f[n][1]);}

"Simulation competition 20181025": Game Theory + dp simple Optimization