BZOJ 1112 POI2008 砖块Klo Treap

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题目大意：给定一个长度为n的序列，求一个长度为k的子区间，将这个长度为k的区间变成一样的，代价总和最小，求最小花销

显然选取的是这k个数的中位数时代价总和最小

于是我们从左往右扫一遍用一个Treap来维护这个长度为k的区间即可

时间复杂度O(nlogn) 这水题居然还贡献了一个WA真是。。。

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define M 100100#define SIZE(p) ((p)?(p)->size:0)#define SUM(p) ((p)?(p)->sum:0)using namespace std;struct Treap{Treap *ls,*rs;int val,key;int cnt,size;long long sum;void* operator new (size_t,int _){static Treap mempool[M],*C=mempool;C->ls=C->rs=0x0;C->sum=C->val=_;C->key=rand();C->cnt=C->size=1;return C++;}void Push_Up(){sum=(long long)val*cnt;size=cnt;if(ls) sum+=ls->sum,size+=ls->size;if(rs) sum+=rs->sum,size+=rs->size;}friend void Zig(Treap *&x){Treap *y=x->ls;x->ls=y->rs;y->rs=x;x=y;x->Push_Up();x->rs->Push_Up();}friend void Zag(Treap *&x){Treap *y=x->rs;x->rs=y->ls;y->ls=x;x=y;x->Push_Up();x->ls->Push_Up();}friend void Insert(Treap *&x,int y){if(!x){x=new (y)Treap;return ;}if(y==x->val)x->cnt++;else if(y<x->val){Insert(x->ls,y);if(x->ls->key>x->key)Zig(x);}else{Insert(x->rs,y);if(x->rs->key>x->key)Zag(x);}x->Push_Up();}friend void Delete(Treap *&x,int y){if(y<x->val)Delete(x->ls,y);else if(y>x->val)Delete(x->rs,y);else if(x->cnt>=2)--x->cnt;else if(!x->ls)x=x->rs;else if(!x->rs)x=x->ls;else{Zag(x);Delete(x->ls,y);if(x->ls&&x->ls->key>x->key)Zig(x);}if(x) x->Push_Up();}friend int Get_Kth(Treap *x,int y){if(y<=SIZE(x->ls))return Get_Kth(x->ls,y);if(y<=SIZE(x->ls)+x->cnt)return x->val;elsereturn Get_Kth(x->rs,y-SIZE(x->ls)-x->cnt);}friend long long Query(Treap *x,int y){if(!x) return 0;if(y<=SIZE(x->ls))return Query(x->ls,y);if(y<=SIZE(x->ls)+x->cnt)return SUM(x->ls) + (long long)x->val*(y-SIZE(x->ls));elsereturn SUM(x->ls) + (long long)x->val*x->cnt + Query(x->rs,y-SIZE(x->ls)-x->cnt);}}*root;int n,k,a[M];long long ans=0x3f3f3f3f3f3f3f3fll,sum[M];int main(){int i;srand(19980402);cin>>n>>k;for(i=1;i<=n;i++){scanf("%d",&a[i]);sum[i]=sum[i-1]+a[i];}for(i=1;i<k;i++)Insert(root,a[i]);for(i=k;i<=n;i++){Insert(root,a[i]);long long mid=Get_Kth(root,k+1>>1);long long lesser=Query(root,k+1>>1);long long greater=sum[i]-sum[i-k]-lesser;ans=min(ans, mid*(k+1>>1)-lesser + greater-mid*(k-(k+1>>1)) );Delete(root,a[i-k+1]);}cout<<ans<<endl;return 0;}

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