BZOJ-2732 [HNOI2012] Archery

It is still turned into an inequality group, and then the half plane is cross.

#include <cstdlib> #include <cstdio> #include <cmath> #include <cstring> #include <cctype > #include <algorithm> #define REP (i, L, R) for (int i=l; i<=r; i++) #define CLR (x, C) memset (x, C, sizeof (x)) #def Ine MAXN 234567#define Double long double#define linf 1e15using namespace std;typedef long long ll;inline int read () {int x =0, f=1; Char Ch=getchar (), while (!isdigit (CH)) {if (ch== '-') f=-1; Ch=getchar ();} while (IsDigit (CH)) x=x*10+ch-' 0 ', Ch=getchar (); return x*f;} struct p{double x, y;}; struct line{p A, b; int id; double ang;} L[MAXN], A[MAXN], Q[MAXN]; P operator-(p a, p B) {return (p) {a.x-b.x, a.y-b.y};} Double operator * (p a, p b) {return a.x*b.y-a.y*b.x;} BOOL Operator < (line A, line B) {if (A.ang==b.ang) return (A.B-A.A) * (B.B-B.A) >0; return A.ang<b.ang;} int n, m, L, R, Cnt;inline double cal (double A, double b, int x) {return b/a-a*x;} P Inter (line A, line B) {double k1= (b.b-a.a) * (A.B-A.A), k2= (A.B-A.A) * (B.A-A.A), t=k2/(K1+K2); return (P) {b.a.x+t* (b.b.x-b.a.x), b.a.y+t* (B.B.Y-B.A.Y)};} BOOL Jud (line A, line B, line t) {return (Inter (A, b)-t.a) * (T.B-T.A) >0;} void HPI (int x) {int Cnt=0;rep (i, 1, m) if (l[i].id<=x) {if (L[i].ang!=a[cnt].ang) cnt++;a[cnt]=l[i];} L=1, r=0;q[++r]=a[1]; Q[++r]=a[2];rep (i, 3, CNT) {while (L<r && Jud (Q[r-1], q[r], a[i]) r--;while (l<r && Jud (q[l+1], q[l) , A[i])) l++;q[++r]=a[i];} while (L<r && Jud (Q[r-1], q[r], q[l]) r--;while (l<r && Jud (q[l+1], q[l], Q[r])) l++;} int main () {n=read (); l[++m].a= (p) {-linf,-linf}; l[m].b= (p) {linf,-linf};l[++m].a= (p) {linf,-linf}; l[m].b= (p) {Linf, Linf};l[++m].a= (P) {linf, linf}; L[m].b= (p) {-linf, linf};l[++m].a= (p) {-linf, linf}; L[m].b= (P) {-linf,-linf};rep (i, 1, n) {double x=read (), Ya=read (), Yb=read (); l[++m].a.x=-1; l[m].a.y=cal (x, ya,-1); L[m] . b.x=1; l[m].b.y=cal (x, ya, 1); l[++m].a.x=1; L[m].a.y=cal (x, YB, 1); L[m].b.x=-1; L[m].b.y=cal (x, YB,-1); l[m].id=l[m-1].id=i;} Rep (i, 1, m) l[i].ang=atan2 (L[I].B.Y-L[I].A.Y, L[I].B.X-L[I].A. x); Sort (l+1, l+m+1); int l=0, R=n;while (l<r) {int mid= (l+r) >>1;hpi (mid+1), if (r-l>1) l=mid+1; else r=mid;} printf ("%d\n", l); return 0;}

BZOJ-2732 [HNOI2012] Archery