【bzoj4293】【PA2015】Siano

如题，首先可以考虑相对大小是不变的。

那么先sort，之后每次在线段树上二分即可。

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
const int N=;
int n,m;
ll a[N],d[N],b[N];
struct Segment_Tree{
#define lson (o<<1)
#define rson (o<<1|1)
    ll addv[N<<],sumv[N<<],suma[N<<],maxv[N<<],setv[N<<];
    inline void pushup(int o){
        sumv[o]=sumv[lson]+sumv[rson];
        maxv[o]=max(maxv[lson],maxv[rson]);
    }
    inline void pushdown(int o,int l,int r){
        int mid=(l+r)>>;
        if(setv[o]!=-){
            maxv[lson]=setv[o];maxv[rson]=setv[o];
            setv[lson]=setv[o];setv[rson]=setv[o];
            sumv[lson]=1LL*setv[o]*(mid-l+);
            sumv[rson]=1LL*setv[o]*(r-mid);
            addv[lson]=;addv[rson]=;setv[o]=-;
        }
        if(addv[o]){
            sumv[lson]+=addv[o]*suma[lson];sumv[rson]+=addv[o]*suma[rson];
            maxv[lson]+=1LL*a[mid]*addv[o];maxv[rson]+=1LL*a[r]*addv[o];
            addv[lson]+=addv[o];addv[rson]+=addv[o];
            addv[o]=;
        }
    }
    inline void build(int o,int l,int r){
        setv[o]=-;
        if(l==r){suma[o]=a[l];return;}
        int mid=(l+r)>>;
        build(lson,l,mid);build(rson,mid+,r);
        suma[o]=suma[lson]+suma[rson];
    }
    ll find(int o,int l,int r,ll v){
        if(l==r){if(maxv[o]>=v)return l;else return -;}
        pushdown(o,l,r);
        int mid=(l+r)>>;
        if(maxv[lson]>=v)return find(lson,l,mid,v);else return find(rson,mid+,r,v);
    }
    ll querysum(int o,int l,int r,int ql,int qr){
        if(ql<=l&&r<=qr)return sumv[o];
        pushdown(o,l,r);int mid=(l+r)>>;;ll ans=;
        if(ql<=mid)ans+=querysum(lson,l,mid,ql,qr);
        if(qr>mid)ans+=querysum(rson,mid+,r,ql,qr);
        return ans;
    }
    inline void change(int o,int l,int r,int ql,int qr,ll v){
        if(ql<=l&&r<=qr){
            maxv[o]=setv[o]=v;sumv[o]=1LL*v*(r-l+);
            addv[o]=;return;
        }
        pushdown(o,l,r);int mid=(l+r)>>;
        if(ql<=mid)change(lson,l,mid,ql,qr,v);
        if(qr>mid)change(rson,mid+,r,ql,qr,v);
        pushup(o);
    }
}T;
inline ll read(){
    ll f=,x=;char ch;
    do{ch=getchar();if(ch=='-')f=-;}while(ch<''||ch>'');
    do{x=x*+ch-'';ch=getchar();}while(ch>=''&&ch<='');
    return f*x;
}
int main(){
    n=read();m=read();
    for(int i=;i<=n;i++)a[i]=read();
    sort(a+,a+n+);
    T.build(,,n);
    for(int i=;i<=m;i++){
        d[i]=read();b[i]=read();
        T.sumv[]+=T.suma[]*(d[i]-d[i-]);T.maxv[]+=1LL*a[n]*(d[i]-d[i-]);
        T.addv[]=d[i]-d[i-];ll t=T.find(,,n,b[i]);
        if(t==-){puts("");continue;}
        ll ans1=T.querysum(,,n,t,n);T.change(,,n,t,n,b[i]);
        ll ans2=T.querysum(,,n,t,n);printf("%lld\n",ans1-ans2);
    }
}