lg4169

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链接:lg4169

题解

当满足的点在询问点左下角时去掉绝对值,转换为三维偏序问题,坐标翻转即可将另三个区域的点转换为左下角的点
或4次cdq采用不同的树状数组最后取最值

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#include<bits/stdc++.h>
#define rep(i,x,y) for(register int i=x;i<=y;++i)
#define lowbit(x) x&(-x)
using namespace std;
const int INF=0x3f3f3f3f;
const int N=1e6+7;
struct node{
    int x,y,t,key,op;
}p[N],t[N],a[N];
int tr[N];
int n,m,mx,my;
inline void update(int x,int y){for(;x<=my;x+=lowbit(x))tr[x]=max(tr[x],y);}
inline int query(int x){int res=-1;for(;x;x-=lowbit(x))res=max(res,tr[x]);return res;}
inline int cmpx(node x,node y){return x.x==y.x?x.y<y.y:x.x<y.x;}
void cdq(int l,int r){
    if(l==r)return ;
    int mid=l+r>>1;
    cdq(l,mid);cdq(mid+1,r);
    int posa=l,posb=mid+1;
    while(posb<=r){
        while(posa<=mid&&a[posa].x<=a[posb].x){
        	if(a[posa].op==1)update(a[posa].y,a[posa].x+a[posa].y);
			++posa;
        }
        if(a[posb].op==0){
            int tp=query(a[posb].y);
            if(tp)p[a[posb].t].key=min(p[a[posb].t].key,a[posb].x+a[posb].y-tp);
        }
        ++posb;
    }
    rep(i,l,posa-1)if(a[i].op==1)for(int j=a[i].y;j<=my;j+=lowbit(j))tr[j]=0;
    //sort(a+l,a+r+1,cmpx);
    int i=l,j=mid+1,k=l-1;
    while(j<=r){
        while(i<=mid&&cmpx(a[i],a[j]))t[++k]=a[i++];
        t[++k]=a[j++];
    }
    while(j<=r)t[++k]=a[j++];
    while(i<=mid)t[++k]=a[i++];
    rep(i,l,r)a[i]=t[i];
}
inline void init(){
    rep(i,1,n)a[i]=p[i];
}
int main(){
    scanf("%d%d",&n,&m);
    rep(i,1,n){
        int x,y;
        scanf("%d%d",&x,&y);
        x++,y++;
        p[i]=node{x,y,i,x+y,1};
        mx=max(x,mx);
        my=max(y,my);
    }
    while(m--){
        int op,x,y;
        scanf("%d%d%d",&op,&x,&y);
        x++,y++;
        if(op==1)p[++n]=node{x,y,n,x+y,1};
        else p[++n]=node{x,y,n,INF,0};
        mx=max(x,mx);
        my=max(y,my);
    }
    init();
    cdq(1,n);
    rep(i,1,n)p[i].x=mx-p[i].x+1;
    init();
    cdq(1,n);
    rep(i,1,n)p[i].y=my-p[i].y+1;
    init();
    cdq(1,n);
    rep(i,1,n)p[i].x=mx-p[i].x+1;
    init();
    cdq(1,n);
    rep(i,1,n)if(p[i].op==0)printf("%d\n",p[i].key);
    return 0;
}