Sol

一个很显然的暴力，设\(f[i]\)表示选到\(i\)的最优效率

每次枚举一段不与前面连续的长度小于\(k\)的区间转移来

# include <bits/stdc++.h>

# define RG register

# define IL inline

# define Fill(a, b) memset(a, b, sizeof(a))

using namespace std;

typedef long long ll;

const int _(1e5 + 5);

IL int Input(){

    RG int x = 0, z = 1; RG char c = getchar();

    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;

    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);

    return x * z;

}

int n, k;

ll f[_], sum[_];

int main(RG int argc, RG char* argv[]){

    n = Input(); k = Input();

    for(RG int i = 1; i <= n; ++i) f[i] = Input(), sum[i] = sum[i - 1] + f[i];

    for(RG int i = 2; i <= n; ++i)

        for(RG int j = max(0, i - k); j < i; ++j)

            f[i] = max(f[i], f[j - 1] + sum[i] - sum[j]);

    printf("%lld\n", f[n]);

    return 0;

}

把转移中的\(f[j-1]\)和\(sum[j]\)写在一起就可以单调队列优化

# include <bits/stdc++.h>

# define RG register

# define IL inline

# define Fill(a, b) memset(a, b, sizeof(a))

using namespace std;

typedef long long ll;

const int _(1e5 + 5);

IL int Input(){

    RG int x = 0, z = 1; RG char c = getchar();

    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;

    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);

    return x * z;

}

int n, k;

ll f[_], g[_], sum[_], Q[_], head, tail = 1;

int main(RG int argc, RG char* argv[]){

    n = Input(); k = Input();

    for(RG int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + Input();

    Q[0] = -1; g[0] -= sum[1];

    for(RG int i = 1; i <= n; ++i){

        while(i - Q[head] - 1 > k) ++head;

        f[i] = (Q[head] == -1 ? 0 : g[Q[head]]) + sum[i], g[i] = f[i] - sum[i + 1];

        while(head <= tail && g[Q[tail]] < g[i]) --tail;

        Q[++tail] = i;

    }

    printf("%lld\n", f[n]);

    return 0;

}