COJ969 WZJ data structure (minus 31)

WZJ data structure (minus 31)

Difficulty level: D; operating time limit: 3000ms; operating space limit: 262144KB; code length limit: 2000000B

Question Description

A state-owned two main base stations, supply the national resources. Defines a primary base station that can cover a circle with a radius of R (including the boundary) at the center of the main base station. If a sub-base station can be overwritten by a primary base station, it is active. There are n events, and there are two types of events: 1. Create a new sub-base station with coordinates (x, y). 2. Give the radius of two main base stations: R1, R2. Inquire about the number of sub-base stations that are inactive.

Input

The first line is four positive integers: x1, y1, x2, y2. Represents the coordinates of the two primary base stations (X1,Y1) and (X2,y2). The second line is a positive integer n, which indicates that there are N events. The next n rows, three positive integers per line. If the first number is 1, then the next two digits are X, Y, which represents a new base station with coordinates (x, y). If the first number is 2, then the next two numbers are R1,R2, which indicates the number of sub-base stations in the inactive state when two primary base stations have a coverage radius of R1 and R2.

Output

Output answers for each query.

Input example

1 10 5 2 10 1 2 6 1 1 9 1 3 8 1 6 7 1 4 12 2 1 1 2 3 2 2 8 2 2 2 2 2 3 2

Output example

4 3 0 4 3

Other Notes

1<=x1,y1,x2,y2,x,y,r1,r2<=10^9 1<=n<=200000

Mom, I'm going to write a line-of-tree Division!

According to the method of the paper, the operation corresponds to a time interval, marking the interval, then the number of marks of the operation is O (NLOGN), the query corresponds to a point, the point is located in the interval marked, then the number of tags asked is O (Nlogn). At last, you can do it offline for each line segment.

#include <cstdio>#include<cctype>#include<queue>#include<cstring>#include<algorithm>#defineRep (i,s,t) for (int i=s;i<=t;i++)#defineDwn (i,s,t) for (int i=s;i>=t;i--)#defineren for (int i=first[x];i!=-1;i=next[i])using namespaceStd;inlineintRead () {intx=0, f=1;CharC=GetChar ();  for(;! IsDigit (c); C=getchar ())if(c=='-') f=-1;  for(; IsDigit (c); C=getchar ()) x=x*Ten+c-'0'; returnx*F;} typedefLong Longll;Const intmaxn=200010; ll X0,Y0,X1,Y1,TMP[MAXN];intn,n,first1[maxn*3],next1[maxn* the],id1[maxn* the],tot1;intfirst2[maxn*3],next2[maxn* the],id2[maxn* the],tot2;structQuery {intTp,id; ll x, y;} Q[MAXN];structPoint {ll x, y;intID; BOOL operator< (Constpoint& a)Const {        if(x!=a.x)returnX>a.x; if(Y!=A.Y)returnY>a.y; }}A[MAXN];voidPre () {Rep (I,1, N) tmp[i]=-q[i].y; Sort (tmp+1, tmp+n+1); Rep (I,1, N) q[i].y=lower_bound (tmp+1, tmp+n+1,-Q[I].Y)-tmp;}intANS[MAXN],SUMV[MAXN];voidAddintXintV) { for(; x<=n;x+=x&-x) sumv[x]+=v;}intSumintx) {intres=0; for(; x;x-=x&-x) res+=sumv[x];returnRes;}voidQuery1 (intOintLintRintQlintQrintv) {if(ql<=l&&r<=qr) {id1[++tot1]=v;next1[tot1]=first1[o];first1[o]=ToT1; }    Else {        intMid=l+r>>1,lc=o<<1, rc=lc|1; if(ql<=mid) Query1 (LC,L,MID,QL,QR,V); if(Qr>mid) Query1 (rc,mid+1, r,ql,qr,v); }} voidQuery2 (intOintLintRintXintv) {id2[++tot2]=v;next2[tot2]=first2[o];first2[o]=ToT2; if(L==R)return; intMid=l+r>>1,lc=o<<1, rc=lc|1; if(x<=mid) Query2 (LC,L,MID,X,V); ElseQuery2 (rc,mid+1, r,x,v);}intMain () {x0=read (); Y0=read (); X1=read (); y1=read (); N=read (); Rep (I,1, N) {Q[q[i].id=i].tp=read (); ll x=read (), y=read (); if(q[i].tp==1) {q[i].x= (x-x0) * (x-x0) + (y-y0) * (y-y0), q[i].y= (x-x1) * (x-x1) + (y-y1) * (yy1); Query1 (1,1, N,i,n,i); }        Else{q[i].x=x*x,q[i].y=y*y; Query2 (1,1, N,i,i);    }} pre (); Rep (I,1,3*N) {intm=0;  for(intJ=FIRST2[I];J;J=NEXT2[J]) a[++m]=(point) {q[id2[j]].x,q[id2[j]].y,q[id2[j]].id};  for(intJ=FIRST1[I];J;J=NEXT1[J]) a[++m]= (point) {q[id1[j]].x,q[id1[j]].y,0}; Sort (A+1, a+m+1); Rep (J,1, m)if(! a[j].id) Add (A[J].Y,1);ElseAns[a[j].id]+=sum (a[j].y-1); Rep (J,1, m)if(! a[j].id) Add (a[j].y,-1); } Rep (I,1, N)if(q[i].tp==2) printf ("%d\n", Ans[i]); return 0;}

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CDQ Divide the constant is relatively small:

#include <cstdio>#include<cctype>#include<queue>#include<cstring>#include<algorithm>#defineRep (i,s,t) for (int i=s;i<=t;i++)#defineDwn (i,s,t) for (int i=s;i>=t;i--)#defineren for (int i=first[x];i!=-1;i=next[i])using namespaceStd;inlineintRead () {intx=0, f=1;CharC=GetChar ();  for(;! IsDigit (c); C=getchar ())if(c=='-') f=-1;  for(; IsDigit (c); C=getchar ()) x=x*Ten+c-'0'; returnx*F;} typedefLong Longll;Const intmaxn=200010; ll X0,Y0,X1,Y1,TMP[MAXN];structQuery {intTp,id; ll x, y;} Q[MAXN];structPoint {ll x, y;intID; BOOL operator< (Constpoint& a)Const {        if(x!=a.x)returnX>a.x; if(Y!=A.Y)returnY>a.y; }}A[MAXN];intN,ANS[MAXN],SUMV[MAXN];voidAddintXintV) { for(; x<=n;x+=x&-x) sumv[x]+=v;}intSumintx) {intres=0; for(; x;x-=x&-x) res+=sumv[x];returnRes;}voidSolveintLintR) {if(L&GT;=R)return; intMid=l+r>>1, m0=0, m=0; Solve (l,mid); Solve (mid+1, R); Rep (I,l,mid)if(q[i].tp==1) a[++m0]= (point) {q[i].x,q[i].y,0}; if(!M0)return; m=M0; Rep (I,mid+1Rif(q[i].tp==2) a[++m0]=(point) {q[i].x,q[i].y,q[i].id}; if(m0==m)return; m=M0; Sort (A+1, a+m+1); Rep (I,1, m)if(! a[i].id) Add (A[I].Y,1);ElseAns[a[i].id]+=sum (a[i].y-1); Rep (I,1, m)if(! a[i].id) Add (a[i].y,-1);}voidPre () {Rep (I,1, N) tmp[i]=-q[i].y; Sort (tmp+1, tmp+n+1); Rep (I,1, N) q[i].y=lower_bound (tmp+1, tmp+n+1,-Q[I].Y)-tmp;}intMain () {x0=read (); Y0=read (); X1=read (); y1=read (); N=read (); Rep (I,1, N) {Q[q[i].id=i].tp=read (); ll x=read (), y=read (); if(q[i].tp==1) q[i].x= (x-x0) * (x-x0) + (y-y0) * (y-y0), q[i].y= (x-x1) * (x-x1) + (y-y1) * (yy1); Elseq[i].x=x*x,q[i].y=y*y; } pre (); Solve (1, N); Rep (I,1, N)if(q[i].tp==2) printf ("%d\n", Ans[i]); return 0;}

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COJ969 WZJ data structure (minus 31)