NOI2010 : 超级钢琴

求出前缀和

对于每个结尾i，设现在取的区间是[j+1,i]，则i-R<=j<=i-L，取出该区间sum[j]的最小值，将sum[i]-sum[j]放入堆中

建立一个大根堆，每次取出堆顶元素，将排名k+1，将sum[i]-区间第k小值放入堆中

求区间第k小可以用主席树

直到取满K次为止

 

#include<cstdio>

#include<algorithm>

#include<queue>

#define N 500010

#define M 10000000

using namespace std;

int n,i,j,K,L,R,sum[N],b[N],l[M],r[M],val[M],root[N],tot;long long ans;

inline void read(int&a){

  char c;bool f=0;a=0;

  while(!((((c=getchar())>='0')&&(c<='9'))||(c=='-')));

  if(c!='-')a=c-'0';else f=1;

  while(((c=getchar())>='0')&&(c<='9'))(a*=10)+=c-'0';

  if(f)a=-a;

}

inline int lower(int x){

  int l=1,r=n,mid,t;

  while(l<=r)if(b[mid=(l+r)>>1]<x)l=mid+1;else r=(t=mid)-1;

  return t;

}

int ins(int x,int a,int b,int c){

  int y=++tot;

  val[y]=val[x]+1;

  if(a==b)return y;

  int mid=(a+b)>>1;

  if(c<=mid)l[y]=ins(l[x],a,mid,c),r[y]=r[x];else l[y]=l[x],r[y]=ins(r[x],mid+1,b,c);

  return y;

}

inline int kth(int x,int y,int k){

  if(val[y]-val[x]<k)return 0;

  int a=1,b=n,mid,t;

  while(1){

    if(a==b)return a;

    t=val[l[y]]-val[l[x]];mid=(a+b)>>1;

    if(k<=t)b=mid,x=l[x],y=l[y];else a=mid+1,k-=t,x=r[x],y=r[y];

  }

}

struct node{

  int x,l,r,k,i;

  node(){}

  node(int _i,int _l,int _r,int _k){i=_i;l=_l;r=_r;k=_k;x=sum[i]-b[kth(root[l],root[r],k)];}

  node(int _i,int _k){i=_i;l=max(i-R-1,0);r=max(i-L,0);k=_k;x=sum[i]-b[kth(root[l],root[r],k)];}

  bool operator<(const node&b)const{return x<b.x;}

}tmp;

priority_queue<node>H;

int main(){

  for(read(n),read(K),read(L),read(R),n++,i=2;i<=n;i++)read(j),sum[i]=sum[i-1]+j;

  for(b[0]=600000000,i=1;i<=n;i++)b[i]=sum[i];

  for(sort(b+1,b+n+1),i=1;i<=n;i++)root[i]=ins(root[i-1],1,n,lower(sum[i]));

  for(i=2;i<=n;i++)H.push(node(i,1));

  while(K--)tmp=H.top(),H.pop(),ans+=tmp.x,H.push(node(tmp.i,tmp.l,tmp.r,tmp.k+1));

  printf("%lld",ans);

  return 0;

}