需要考虑的形式是 $-b+c$,所以元素可以分为两类,一种是前面带负号的,例如 $b$,另一种是前面带正号的,例如 $c$。
$d[i][0]$ 表示前 $i$ 个元素中,以 $c$ 类元素结尾的最大价值。$d[i][1]$ 表示前 $i$ 个元素中,以 $b$ 类元素结尾的最大价值。有了这两个状态,那么状态转移方程就比较好想了。
// Date: Thu Dec 21 19:53:02 2023
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>
using namespace std;
const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
{0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;
const ull Pr = 131;
#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])
#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())
template <typename t> istream &operator>>(istream &in, vector<t> &vec) {
for (t &x : vec)
in >> x;
return in;
}
template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) {
int n = SZ(vec);
For(i, 0, n) {
out << vec[i];
if (i < n - 1)
out << ' ';
}
return out;
}
#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z) \
cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif
const int N = 300010;
int t, n, q, a[N];
ll d[N][2];
int main(void) {
#ifdef _DEBUG
freopen("1420c1.in", "r", stdin);
#endif
std::ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> t;
while (t--) {
cin >> n >> q;
memset(d, 0, sizeof d);
For1(i, 1, n) {
cin >> a[i];
d[i][0] = a[i];
}
For1(i, 2, n) {
d[i][1] = max(d[i - 1][0] - a[i], d[i - 1][1]);
d[i][0] = max(d[i][0], max(d[i - 1][1] + a[i], d[i - 1][0]));
}
cout << max(d[n][0], d[n][1]) << '\n';
}
return 0;
}