我们很容易想到nk的做法:
定义f[i]为前i个数这样分的方法
那么转移为
改成前缀和
若一个点比另一个优,(设k比j优),那么f[k - 2] + sum[i] - sum[k - 1] > f[j - 2] + sum[i] - sum[j - 1], 即f[k - 2] - sum[k - 1] > f[j - 2] - sum[j - 1]。我们用单调队列维护就好
一开始把while写成if调了半天,QAQ
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cmath>
const int N = 1e5 + 10;
int n, k, head = 1, tail = 1;
long long sum[N], f[N];
struct item {
int id;
long long data;
item () {}
item (int id, long long data) : id (id), data (data) {}
bool operator >= (const item &rhs) const {
return data >= rhs.data;
}
}q[N], hel;
int main () {
scanf ("%d%d", &n, &k);
q[1] = item (1, 0);
for (int i = 1; i <= n; ++i) {
scanf ("%lld", sum + i); sum[i] += sum[i - 1];
if (i >= 2) {
hel = item (i, f[i - 2] - sum[i - 1]);
while (head <= tail && hel >= q[tail]) --tail;
q[++tail] = hel;
}
while (head <= tail && i - k + 1 > q[head].id) ++head;
f[i] = std :: max (q[head].data + sum[i], f[i - 1]);
}
printf ("%lld
", f[n]);
return 0;
}