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;
}