bzoj2442(单调队列优化)

来源：互联网发布：mac单引号怎么打出来编辑：程序博客网时间：2024/05/01 06:01

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f[i]=max(f[j−2]+sum[j][i]) (i−k+1<=j<=i)
改成前缀和f[i]=max(f[j−2]+sum[i]−sum[j−1]) (i−k+1<=j<=i)
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\n", f[n]);    return 0;}

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