Poj 3335 rotating scoreboard

Same as poj3130

Solution: Same as poj3130

Poj3130 solution Report: click here

Note: The order of points given by the two questions is different. Do not copy templates all the time (I won't tell you that I did it ).

The following code is used:

# Include <set> # include <map> # include <queue> # include <math. h> # include <vector> # include <string> # include <stdio. h> # include <string. h> # include <stdlib. h >#include <iostream> # include <algorithm> # define EPS 1e-8 # define PI ACOs (-1.0) # define INF 107374182 # define inf64 1152921504606846976 # define LC l, m, tr <1 # define RC m + 1, R, TR <1 | 1 # define zero (a) FABS (a) <EPS # define iabs (X) (x)> 0? (X):-(x) # define clear1 (A, X, size) memset (A, X, sizeof (A [0]) * (size )) # define clearall (A, x) memset (A, X, sizeof (A) # define memcopy1 (A, X, size) memcpy (A, X, sizeof (X [0]) * (size) # define memcopyall (A, x) memcpy (A, X, sizeof (x) # define max (x, y) (x)> (y ))? (X): (y) # define min (x, y) (x) <(y ))? (X): (y) using namespace STD; const int maxn = 55; int DQ [maxn], top, bot, Pn, order [maxn], ln; struct point {Double X, Y;} p [maxn]; struct line {point a, B; double angle;} l [maxn]; int dblcmp (Double K) {If (FABS (k) <EPS) return 0; return K> 0? 1:-1;} Double Multi (point P0, Point P1, point P2) {return (p1.x-Snapshot X) * (p2.y-Snapshot y)-(p1.y-Snapshot y) * (p2.x-0000x);} bool CMP (int u, int v) {int d = dblcmp (L [u]. angle-L [v]. angle); If (! D) return dblcmp (multi (L [u]. a, L [v]. a, L [v]. b) <0; // the left half of the orientation is divided into half plane, less than 0. Return d <0;} void getintersect (line L1, line L2, point & P) {double dot1, dot2; dot1 = multi (l2.a, l1. B, l1.a); dot2 = multi (l1. B, l2. B, l1.a); p. X = (l2.a. x * dot2 + l2. B. x * dot1)/(dot2 + dot1); p. y = (l2.a. y * dot2 + l2. B. y * dot1)/(dot2 + dot1);} bool judge (line l0, line L1, line l2) {point P; getintersect (L1, L2, P ); return dblcmp (multi (p, l0.a, l0. B)> 0; // The less than symbol is opposite to the comment in CMP () above} void addline (double X1, double Y1, double X2, double Y2) {L [Ln]. a. X = x1; L [Ln]. a. y = Y1; L [Ln]. b. X = x2; L [Ln]. b. y = Y2; L [Ln]. angle = atan2 (y2-y1, x2-x1); Order [Ln] = ln; Ln ++;} void halfplaneintersection () {int I, j; sort (Order, order + ln, CMP); for (I = 1, j = 0; I <ln; I ++) if (dblcmp (L [order [I]. angle-L [order [J]. angle)> 0) Order [+ + J] = order [I]; Ln = J + 1; DQ [0] = order [0]; DQ [1] = order [1]; bot = 0; Top = 1; for (I = 2; I <ln; I ++) {While (BOT <top & judge (L [order [I], L [DQ [Top-1], L [DQ [Top]) Top --; while (BOT <top & judge (L [order [I], L [DQ [bot + 1], L [DQ [bot]) BOT ++; DQ [++ top] = order [I];} while (BOT <top & judge (L [DQ [bot], L [DQ [Top-1], L [DQ [Top]) Top --; while (BOT <top & judge (L [DQ [Top], L [DQ [bot + 1], L [DQ [bot]) BOT ++;} bool isthereacore () {If (top-bot> 1) return true; return false;} int main () {int I, T; scanf ("% d", & T); While (t --) {scanf ("% d", & PN); for (I = 0; I <PN; I ++) scanf ("% lf", & P [I]. x, & P [I]. y); For (Ln = I = 0; I <pn-1; I ++) addline (P [I]. x, p [I]. y, P [I + 1]. x, p [I + 1]. y); addline (P [I]. x, p [I]. y, P [0]. x, p [0]. y); halfplaneintersection (); If (isthereacore () printf ("Yes \ n"); else printf ("NO \ n");} return 0 ;}

Poj 3335 rotating scoreboard