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【wqs二分决策单调性】HHHOJ#261. Brew

程序员亮亮技术 2022年11月15日

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第一道决策单调性……

题目描述

HHHOJ#261. Brew

题目分析

挺好的……模板题？

寄存了先。

 #include<bits/stdc++.h>
 typedef long long ll;
 const int maxn = ; struct node
 {
     int x,l,r;
     node(int a=, int b=, int c=):x(a),l(b),r(c) {}
 };
 int n,k,a[maxn],g[maxn];
 ll s[maxn],mid,pos,ans,f[maxn];
 std::deque<node> q; int read()
 {
     char ch = getchar();
     int num = ;
     bool fl = ;
     for (; !isdigit(ch); ch=getchar())
         if (ch=='-') fl = ;
     for (; isdigit(ch); ch=getchar())
         num = (num<<)+(num<<)+ch-;
     if (fl) num = -num;
     return num;
 }
 void clears(std::deque<node> &q)
 {
     std::deque<node> emt;
     std::swap(q, emt);
 }
 ll val(int l, int r)
 {
     if (l >= r) return ;
     int mid = (l+r)>>;
     return 1ll*(mid-l+)*a[mid]-(s[mid]-s[l-])+(s[r]-s[mid-])-1ll*(r-mid+)*a[mid];
 }
 ll calc(int l, int r)
 {
     return f[l]+val(l+, r);
 }
 bool check(ll w)
 {
     clears(q), q.push_front(node(, , n));
     for (int i=; i<=n; i++)
     {
 //        printf("w:%lld i:%d\n",w,i);
         node tt = q.front();
         g[i] = g[tt.x]+, f[i] = calc(tt.x, i)+w;
         int upd = -;
         while (q.size())
         {
             node tt = q.back();
             if (calc(tt.x, tt.l) > calc(i, tt.l))
                 upd = tt.l, q.pop_back();
             else{
                 int lbd = tt.l, rbd = tt.r, fnd = -;
                 while (lbd <= rbd)
                 {
                     int mid = (lbd+rbd)>>;
                     if (calc(tt.x, mid) > calc(i, mid))
                         fnd = mid, rbd = mid-;
                     else lbd = mid+;
                 }
                 if (fnd!=-) upd = fnd, q.back().r = fnd-;
                 break;
             }
         }
         if (upd!=-) q.push_back(node(i, upd, n));
         if (q.size()){
             q.front().l++;
             if (q.front().l > q.front().r) q.pop_front();
         }
     }
     return g[n] <= k;
 }
 int main()
 {
     n = read(), k = read();
     if (k >= n) puts("");
     else{
         for (int i=; i<=n; i++) a[i] = read();
         std::sort(a+, a+n+);
         for (int i=; i<=n; i++) s[i] = s[i-]+a[i];
         ll l = , r = 1e12;
         for (; l<=r; )
         {
             mid = (l+r)>>;
             if (check(mid))
                 ans = f[n]-mid*k, r = mid-;
             else l = mid+;
         }
         printf("%lld\n",ans);
     }
     return ;
 }