[Simulated annealing two-point answer] Bzoj 1038 [ZJOI2008] Observation tower _ two points

Positive solution half plane intersection,%%% popoqqq:http://blog.csdn.net/popoqqq/article/details/39340759

"To determine the height of the Watchtower, we choose the binary processing for each of the two points. We enumerate the endpoints on the polyline from left to right. The line to each endpoint must be counterclockwise from left to right, or it will be obscured.

#include <cstdio> #include <cstdlib> #include <algorithm> #include <cmath> #include <ctime

> #define EPS 1e-4 using namespace std;
	Inline char NC () {static Char buf[100000],*p1=buf,*p2=buf;
	if (P1==P2) {p2= (p1=buf) +fread (Buf,1,100000,stdin); if (P1==P2) return EOF;}
return *p1++;
	} inline void read (int &x) {char c=nc (), b=1; for (;!
	(c>= ' 0 ' && c<= ' 9 '); C=nc ()) if (c== '-') b=-1; for (x=0;c>= ' 0 ' && c<= ' 9 '; x=x*10+c-' 0 ', C=NC ());
X*=b;
	} inline int dcmp (double a,double b) {if (Fabs (a-b) <1e-5) return 0;
	if (a<b) return-1;
return 1;
	} struct point{double x,y; Point (Double x=0,double y=0): X (x), Y (y) {} friend double Cross (point p1,point p2,point p0) {return (p1.x-p0.x) * (p2.y-p0
	. Y)-(p2.x-p0.x) * (P1.Y-P0.Y);

}}p[305];
int n;
Double x1,xn;

Double ans=1e130;

Inline double rnd () {return (rand ()%1000)/1000.0;} Inline Double calc (double x) {for (int i=1;i<n;i++) if (dcmp (p[i].x,x) <=0 && dcmp (x,p[i+1].x) <=0 return p[i].y+ (P[I+1].Y-P[I].Y)/(p[i+1].x-p[i].x) * (x-p[i].x);
Point q[305];

int R;
	inline int C (point E) {r=0;
		for (int i=1;i<=n;i++) {while (R && dcmp (Cross (p[i],q[r],e), 0) >=0) r--;
	Q[++r]=p[i];
} return R;  
	} inline bool Check (point E) {for (int i=2;i<=n;i++) if (dcmp (Cross (p[i-1],p[i],e), 0) <0) return 0;
return 1;
	Inline double Bin (double x) {double l=0,r=1e11,mid;
	Double Y=calc (x);
		while (r-l>1e-5) if (Check (mid= (l+r)/2)) R=mid;
	else L=mid;
	Ans=min (ans, (l+r)/2);
Return (L+R)/2;
	inline void SA () {#define BETA. #define EPS 1e-5 Double Nx,sx,now,delta,ret;
	sx= (X1+XN)/2;
	Now=bin (SX); for (double t=xn-x1; t>eps;
		T*=beta) {nx=sx+t* (rnd () *2-1);   
		if (nx<x1 | | nx>xn) continue;
		Ret=bin (NX);
		Delta=now-ret;
	if (delta>0 | | rnd () <exp (DELTA/T)) Sx=nx,now=ret;  
	the int main () {#define CASE 2 srand (10086);
	int IX;
	Freopen ("T.in", "R", stdin); Freopen ("T.out"," W ", stdout);
	Read (n);
	for (int i=1;i<=n;i++) read (ix), P[i].x=ix;
	for (int i=1;i<=n;i++) read (ix), P[i].y=ix; x1=p[1].x;
	xn=p[n].x;
	for (int i=1;i<=case;i++) SA ();
	printf ("%.3lf\n", ans);
return 0; }