[BZOJ2716] [Violet 3] angel doll (CDQ partition)
Face questions
Ayu seven years ago I had received an angel doll, when she took it as a time capsule buried in the ground. Today, seven years later, Ayu forgot her angel doll buried in the where, so she decided only by a hazy memory to find it.
We live in a small town Ayu as a two-dimensional plane coordinate system, while Ayu will remember from time to time may be planted angel dolls at some point (xmy); or Ayu will ask you, if she (x, y), then how far away from the place near her angel doll possible buried.
Because Ayu will act in a direction parallel to the axis, so this question we define the distance between two points is dist (A, B) = | Ax-Bx | + | Ay-By |. Wherein Ax represents the abscissa of the point A, the remaining similar.
analysis
Since the distance from Manhattan, the absolute value of the symbol hard to deal with, we consider the classification discussion.
If the answer to the interrogation point A, the lower left, then the distance \ (x_A-x_B + y_A- y_B = (x_A + y_A) - (x_B + y_B) \)
Since \ (x_A + y_A \) is constant, we want to minimize the distance, as long as a request asking the lower left point A \ ((x_B + y_B) \ ) the minimum point can be. This is a classic divide and conquer cdq problem, this is a classic problem of three-dimensional partial ordering details, please refer to [ BZOJ 2683] simple questions (CDQ partition)
Otherwise how to do it? In fact, we can avoid discussing classified by folding point. If all points in y coordinates of the minimum Ymin, we point to the straight line y = ymin symmetrical, so that the original point A into the top of the original point A, below, can be processed using the above method. x coordinate of empathy. So long as we change the coordinates, and then run CDQ can divide and conquer. They walked a total of four times CDQ partition
Constant larger, often note cards
Code
//分类讨论上下左右四种情况
//通过翻转坐标全部变成左下的情况
//min(x-x'+y-y')=x+y-max(x'+y')
//转化成三维偏序查最大值: x,y,时间都小的最大值
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 2000000
#define INF 0x3f3f3f3f
#define register
using namespace std;
inline void qread(int &x){
x=0;
int sign=1;
char c=getchar();
while(c<'0'||c>'9'){
if(c=='-') sign=-1;
c=getchar();
}
while(c>='0'&&c<='9'){
x=x*10+c-'0';
c=getchar();
}
x=x*sign;
}
inline void qprint(int x){
if(x<0){
putchar('-');
qprint(-x);
}else if(x==0){
putchar('0');
return;
}else{
if(x>=10) qprint(x/10);
putchar(x%10+'0');
}
}
int n,m,cnt,qcnt;
struct node{
int a;
int b;
int c;
int type;
int id;
int ans;
node(){
}
node(int _a,int _b,int _c,int _type){
a=_a;
b=_b;
c=_c;
type=_type;
id=0;
ans=-INF;
}
node(int _a,int _b,int _c,int _type,int _id){
a=_a;
b=_b;
c=_c;
type=_type;
id=_id;
ans=-INF;
}
};
node q[maxn+5],dat[maxn+5];//dat保存翻转前的变量
int cmpa(node p,node q){
if(p.a==q.a){
if(p.b==q.b) return p.c<q.c;
else return p.b<q.b;
}else return p.a<q.a;
}
int cmpb(node p,node q){
if(p.b==q.b) return p.c<q.c;
else return p.b<q.b;
}
struct fenwick_tree{//树状数组可以维护前缀最大值，且常数小
int c[maxn+5];
inline int lowbit(int x){
return x&(-x);
}
inline void update(int pos,int val){
for(int i=pos;i<=cnt;i+=lowbit(i)) c[i]=max(c[i],val);
}
inline void clear(int pos){
for(int i=pos;i<=cnt;i+=lowbit(i)) c[i]=-INF;
}
inline int query(int pos){
int ans=-INF;
for( int i=pos;i>0;i-=lowbit(i)) ans=max(ans,c[i]);
return ans;
}
}T;
node tmp[maxn+5];
void cdq_divide(int l,int r){
if(l==r) return;
int mid=(l+r)>>1;
cdq_divide(l,mid);
cdq_divide(mid+1,r);
int ptr=l-1;
for(int i=mid+1;i<=r;i++){
if(q[i].type==2){//这样写有玄学优化作用，原因未知.
while(ptr<mid&&q[ptr+1].b<=q[i].b){
ptr++;
if(q[ptr].type==1) T.update(q[ptr].c,q[ptr].a+q[ptr].b);
}
q[i].ans=max(q[i].ans,T.query(q[i].c));
}
}
for(int i=l;i<=ptr;i++) if(q[i].type==1) T.clear(q[i].c);
int num=l-1;
int pl=l,pr=mid+1;
while(pl<=mid&&pr<=r){
if(cmpb(q[pl],q[pr])) tmp[++num]=q[pl++];
else tmp[++num]=q[pr++];
}
while(pl<=mid) tmp[++num]=q[pl++];
while(pr<=r) tmp[++num]=q[pr++];
for(int i=l;i<=r;i++) q[i]=tmp[i];
}
int res[maxn+5];
int main(){
int t,x,y;
int maxx,maxy,minx,miny;
maxx=maxy=-INF;
minx=miny=INF;
qread(n);
qread(m);
for(int i=1;i<=n;i++){
qread(x);
qread(y);
cnt++;
q[cnt]=node(x,y,cnt,1);
maxx=max(maxx,x);
minx=min(minx,x);
maxy=max(maxy,y);
miny=min(miny,y);
}
for(int i=1;i<=m;i++){
qread(t);
qread(x);
qread(y);
if(t==1){
cnt++;
q[cnt]=node(x,y,cnt,1);
}else{
cnt++;
qcnt++;
q[cnt]=node(x,y,cnt,2,qcnt);
}
maxx=max(maxx,x);
minx=min(minx,x);
maxy=max(maxy,y);
miny=min(miny,y);
}
for(int i=1;i<=cnt;i++) dat[i]=q[i];
for(int i=1;i<=qcnt;i++) res[i]=INF;
for(int i=1;i<=cnt;i++) T.c[i]=-INF;
sort(q+1,q+1+cnt,cmpa);//左下
cdq_divide(1,cnt);
for(int i=1;i<=cnt;i++){
if(q[i].type==2) res[q[i].id]=min(res[q[i].id],q[i].a+q[i].b-q[i].ans);
}
for(int i=1;i<=cnt;i++){//右下
q[i]=dat[i];
q[i].a=2*maxx-q[i].a;
}
sort(q+1,q+1+cnt,cmpa);
cdq_divide(1,cnt);
for(int i=1;i<=cnt;i++){
if(q[i].type==2) res[q[i].id]=min(res[q[i].id],q[i].a+q[i].b-q[i].ans);
}
for(int i=1;i<=cnt;i++){//左上
q[i]=dat[i];
q[i].b=2*maxy-q[i].b;
}
sort(q+1,q+1+cnt,cmpa);
cdq_divide(1,cnt);
for(int i=1;i<=cnt;i++){
if(q[i].type==2) res[q[i].id]=min(res[q[i].id],q[i].a+q[i].b-q[i].ans);
}
for(int i=1;i<=cnt;i++){//右上
q[i]=dat[i];
q[i].a=2*maxx-q[i].a;
q[i].b=2*miny-q[i].b;
}
sort(q+1,q+1+cnt,cmpa);
cdq_divide(1,cnt);
for(int i=1;i<=cnt;i++){
if(q[i].type==2) res[q[i].id]=min(res[q[i].id],q[i].a+q[i].b-q[i].ans);
}
for(int i=1;i<=qcnt;i++){
qprint(res[i]);
putchar('\n');
}
}