Zoj 3157 weapon Inverse number/tree array

B-weaponTime limit:1000MS Memory Limit:32768KB 64bit IO Format:%lld &%llu Submit Status

Description

In the World War 3, your countries ' scientists has invented a special weapon. Assume that the enemy's city can is described by rectangular coordinates and it has N roads which is all lines . None of the road is paralled with Y-axis. Besides, each road was represented by and different points (ai,bi) (ci,di) on I T. Any three roads won't intersect at one point.

This special weapon can destroy all the castles whose x coordinate belongs to (l,r). After spying, you know the "all the" The castles is set in the "crossing point of" and "roads" and "crossing" is A castle. In addition, each road's end-point ' s X coordinate does not belong to (l,r).

The scientists want to check the weapon ' s effect. If its effect can not reach army ' s expectation, they has to spend more time and more money in expanding its range. Obviously, the number of castles it can destroy plays an important role on the effect. So is asked to calculate how many castles can is destroyed by this special weapon.

Input

Input contains multiple cases.

Every test case, the first line was an integers n (2 <= n <= 10000). then n lines follow. The (i+1)-th line contains four integers ai,bi,ci,di ( -1E8 <= Ai,bi,ci,di <= 1E8). The (n+2)-th line contains-doubles l,r ( -1E8 <= l,r <= 1E8) There is a blank line between Cases.

Output

For each case, the output of the number of castles is destroyed by the weapon.

Sample Input

30 0 1 12 0 1 10 0 2 00 2.5

Sample Output

2

#include <cstdio>#include<cmath>#include<cstring>#include<ctime>#include<iostream>#include<algorithm>#include<Set>#include<vector>#include<sstream>#include<queue>#include<typeinfo>#include<fstream>typedefLong Longll;using namespacestd;//freopen ("d.in", "R", stdin);//freopen ("D.out", "w", stdout);#defineSspeed ios_base::sync_with_stdio (0); Cin.tie (0)#defineMAXN 10001#defineEPS 1e-6Const intinf=0x7fffffff;//infinitely LargeintN;structnode{Doublex; Doubley;};DoubleD[maxn];node Point1[maxn],point2[maxn];node KISS[MAXN],KILL[MAXN];intLowbit (intx) {    returnx& (-x);}voidUpdate1 (intx) {     while(x<=N) {d[x]++; X+=lowbit (x); }}voidUpdate2 (intXintnum) {     while(x>0) {D[x]+=num; X-=lowbit (x); }}intGETSUM1 (intx) {    ints=0;  while(x>0) {s+=D[x]; X-=lowbit (x); }    returns;}BOOLCMP (node X,node y) {if(x.x==y.x)returnx.y<y.y; returnx.x<y.x;}intMain () {sspeed; intN;  while(cin>>N) {memset (d,0,sizeof(d)); N=N;  for(intI=0; i<n;i++) {cin>>point1[i].x>>point1[i].y>>point2[i].x>>point2[i].y; }        DoubleL,r; CIN>>l>>R; L+=EPS; R-=EPS;  for(intI=0; i<n;i++)        {            DoubleK; K= (POINT2[I].Y-POINT1[I].Y)/(point2[i].x-point1[i].x); kill[i].x=k* (l-point2[i].x) +point2[i].y; Kill[i].y=k* (r-point2[i].x) +point2[i].y;//Find out} ll ans=0; Sort (Kill,kill+n,cmp);  for(intI=0; i<n;i++) {kiss[i].x=kill[i].y; Kiss[i].y=i+1; } sort (Kiss,kiss+n,cmp);  for(inti=n-1; i>=0; i--) {ans+=getSum1 (KISS[I].Y);        Update1 (KISS[I].Y); } cout<<ans<<Endl; }    return 0;}

Zoj 3157 weapon Inverse number/tree array