BZOJ5319 JSOI2018列队（主席树）

　　显然集合后相对位置不变最优。主席树上二分向左和向右的分界点即可。注意主席树的值域。我怎么天天就写点一眼题啊。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cassert>
using namespace std;
#define ll long long
#define N 500010
#define V 1500000
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,a[N],root[N],cnt;
struct data{int l,r,x;ll sum;
}tree[N<<5];
void ins(int &k,int l,int r,int x)
{
    tree[++cnt]=tree[k],k=cnt;tree[k].x++;tree[k].sum+=x;
    if (l==r) return;
    int mid=l+r>>1;
    if (x<=mid) ins(tree[k].l,l,mid,x);
    else ins(tree[k].r,mid+1,r,x);
}
int query(int x,int y,int l,int r,int p)
{
    if (l==r) return l;
    int mid=l+r>>1;
    if (p+tree[tree[y].l].x-tree[tree[x].l].x-1<=mid) return query(tree[x].l,tree[y].l,l,mid,p);
    else return query(tree[x].r,tree[y].r,mid+1,r,p+tree[tree[y].l].x-tree[tree[x].l].x);
}
ll sum(int x,int y,int l,int r,int p)
{
    if (l==r) return tree[y].sum-tree[x].sum;
    int mid=l+r>>1;
    if (p<=mid) return sum(tree[x].l,tree[y].l,l,mid,p);
    else return sum(tree[x].r,tree[y].r,mid+1,r,p)+tree[tree[y].l].sum-tree[tree[x].l].sum;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj5319.in","r",stdin);
    freopen("bzoj5319.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read();
    for (int i=1;i<=n;i++) a[i]=read();
    for (int i=1;i<=n;i++)
    {
        root[i]=root[i-1];
        ins(root[i],0,V,a[i]);
    }
    while (m--)
    {
        int l=read(),r=read(),x=read();
        int p=query(root[l-1],root[r],0,V,x),L=x,R=x+r-l;
        ll ans=(1ll*(L+p)*(p-L+1)>>1)-(1ll*(p+1+R)*(R-p)>>1);
        ans+=tree[root[r]].sum-tree[root[l-1]].sum-(sum(root[l-1],root[r],0,V,p)<<1);
        printf(LL,ans);
    }
    return 0;
}