计算几何

转载 于 2018-02-23 16:03:00 发布 · 74 阅读 · 0 ·
CC 4.0 BY-SA版权
原文链接：http://www.cnblogs.com/Serene-shixinyi/p/8462437.html

谢Achen提醒，差点忘了有计算几何这东西。

poj2187 平面最远点对，凸包+旋转卡壳模板

//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
const int maxn=5e4+7;
int n;

char cc; ll ff;
template<typename T>void read(T& aa) {
	aa=0;cc=getchar();ff=1;
	while((cc<'0'||cc>'9')&&cc!='-') cc=getchar();
	if(cc=='-') ff=-1,cc=getchar();
	while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
	aa*=ff;
}

const double eps=1e-10;
double pf(double x){return x*x;}
inline int dcmp(const long double &x){return abs(x)<eps? 0:(x>0? 1:-1);}

struct Xl{
	double x,y;
	Xl(double x=0.,double y=0.):x(x),y(y){}
	bool operator == (const Xl &b) const{return (!dcmp(x-b.x))&&(!dcmp(y-b.y));}
	bool operator <  (const Xl &b) const{return !dcmp(x-b.x)? y<b.y:x<b.x;}
	Xl operator + (const Xl &b) const{return Xl(x+b.x,y+b.y);}
	Xl operator - (const Xl &b) const{return Xl(x-b.x,y-b.y);}
	Xl operator * (const double &b) const{return Xl(x*b,y*b);}
	Xl operator / (const double &b) const{return Xl(x/b,y/b);}
	double len() const{return pf(x)+pf(y);}
}node[maxn],zz[maxn];
int t;

double D_(const Xl& a,const Xl& b){return a.x*b.x+a.y*b.y;}
double X_(const Xl& a,const Xl& b){return a.x*b.y-a.y*b.x;}

bool cmp(const Xl& a,const Xl& b){
	return dcmp(X_(a-node[1],b-node[1]))==0? 
			(a-node[1]).len()<(b-node[1]).len() : dcmp(X_(a-node[1],b-node[1]))>0;
}

void graham() {
	For(i,2,n) if(node[i]<node[1]) swap(node[i],node[1]);
	sort(node+2,node+n+1,cmp);
	zz[++t]=node[1]; zz[++t]=node[2];
	For(i,3,n) {
		while(t>1&&dcmp(X_(zz[t]-zz[t-1],node[i]-zz[t]))<=0) --t;// "<" -> WA
		zz[++t]=node[i];
	}
}

double RC() {
	zz[t+1]=zz[1];
	int now=2;double rs=0;
	For(i,1,t) {
		while(X_(zz[i+1]-zz[i],zz[now]-zz[i])<X_(zz[i+1]-zz[i],zz[now+1]-zz[i]))
			if((++now)>t) now=1;
		rs=max(rs,(zz[now]-zz[i]).len());
	}
	return rs;
}

int main() {
	read(n);
	For(i,1,n) {
		read(node[i].x); 
		read(node[i].y);
	}
	graham();
	printf("%lld",(ll)RC());
	return 0;
}
/*
4
0 0
0 1
1 3
1 0
*/

　　

hdu1007 平面最近点对模板

分治，时间复杂度是nlog(n)的，由鸽巢原理可知，在那个for套for的循环里面枚举实际上最多只会枚举6个点(不会证)

//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long
#define db double
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
const int maxn=1e5+7;
const db INF=1e18;
int n; db ans;

char cc; ll ff;
template<typename T>void read(T& aa) {
	aa=0;cc=getchar();ff=1;
	while((cc<'0'||cc>'9')&&cc!='-') cc=getchar();
	if(cc=='-') ff=-1,cc=getchar();
	while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
	aa*=ff;
}

const db eps=1e-10;
db pf(db x){return x*x;}

struct Xl{
	db x,y;
	Xl(db x=0.,db y=0.):x(x),y(y){}
	Xl operator + (const Xl& b)const{return Xl(x+b.x,y+b.y);}
	Xl operator - (const Xl& b)const{return Xl(x-b.x,y-b.y);}
	db len()const{return sqrt(pf(x)+pf(y));}
}node[maxn],zz[maxn];

bool cmp1(const Xl& a,const Xl& b) {
	return a.x==b.x? a.y<b.y : a.x<b.x;
}

bool cmp2(const Xl& a,const Xl& b) {
	return a.y<b.y;
}

void solve(int l,int r) {
	if(l>=r) return;
	if(l+1==r) {
		ans=min(ans,(node[l]-node[r]).len());
		return;
	}
	int mid=(l+r)>>1,t=0;
	solve(l,mid); solve(mid+1,r);
	For(i,l,r) if(fabs(node[i].x-node[mid].x)<ans) zz[++t]=node[i];
	sort(zz+1,zz+t+1,cmp2);
	For(i,1,t) for(int j=i+1;j<=t&&zz[j].y-zz[i].y<ans;++j) 
		ans=min(ans,(zz[j]-zz[i]).len());
}

int main() {
	read(n);db x,y;
	while(n) {
		For(i,1,n) {
			scanf("%lf%lf",&x,&y);
			node[i]=Xl(x,y);
		}
		sort(node+1,node+n+1,cmp1);
		ans=INF; solve(1,n);
		printf("%.2f\n",ans/2);
		read(n);
	}
	return 0;
}

 

poj3335 判断多边形内核存在 ，半平面交模板

注意onleft中要写>=而不是>这样一个点的情况才能把边保留下来

最后不能也判面积否则就把一个点的情况判掉了。

为此debug了一下午。

//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long
#define For(i,a,b) for(int i=(a);i<=(b);++i)
const int maxn=300+7;
int T,n;

char cc; ll ff;
template<typename T>void read(T& aa) {
	aa=0;cc=getchar();ff=1;
	while((cc<'0'||cc>'9')&&cc!='-') cc=getchar();
	if(cc=='-') ff=-1,cc=getchar();
	while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
	aa*=ff;
}

const double eps=1e-8;
inline int dcmp(const long double &x){return fabs(x)<eps? 0:(x>0? 1:-1);}

struct Xl{
	double x,y;
	Xl(double x=0.,double y=0.):x(x),y(y){}
	Xl operator + (const Xl &b) const{return Xl(x+b.x,y+b.y);}
	Xl operator - (const Xl &b) const{return Xl(x-b.x,y-b.y);}
	Xl operator * (const double &b) const{return Xl(x*b,y*b);}
	double angle() const{return atan2(y,x);} 
}node[maxn],p[maxn];

double D_(const Xl& a,const Xl& b){return a.x*b.x+a.y*b.y;}
double X_(const Xl& a,const Xl& b){return a.x*b.y-a.y*b.x;}

struct Li{
	Xl u,v; double rad;
	Li(Xl u=Xl(0.,0.),Xl v=Xl(0.,0.)):u(u),v(v){rad=(v-u).angle();}
	bool operator < (const Li &b) const{return rad<b.rad;}
}li[maxn],zz[maxn];

Xl lijd(const Li &a,const Li& b) {
	Xl x=a.v-a.u,y=b.v-b.u,z=a.u-b.u;
	long double c=X_(y,z);
	long double t=c/(c+X_(x+z,y));
	return a.u+x*t;
}

bool onleft(const Xl &a,const Li &l) {return dcmp(X_(l.v-l.u,a-l.u))>=0;}//">" -> WA

bool work() {
	For(i,1,n-1) li[i]=Li(node[i],node[i+1]);
	li[n]=Li(node[n],node[1]);
	sort(li+1,li+n+1);
	int s=1,t=0; zz[++t]=li[1];
	For(i,2,n) {
		while(s<t&&!onleft(p[t-1],li[i])) t--;
		while(s<t&&!onleft(p[s],li[i])) s++;
		zz[++t]=li[i];
		if(!dcmp(X_(li[i].v-li[i].u,zz[t-1].v-zz[t-1].u)))
			if(onleft(li[i].u,zz[--t])) zz[t]=li[i];
		if(s<t) p[t-1]=lijd(zz[t],zz[t-1]);
	}
	while(s<t&&!onleft(p[t-1],li[s])) --t;
	return t-s>1;//
}

int main() {
	read(T);
	ll x,y;
	while(T--) {
		read(n);
		For(i,1,n) {
			read(x); read(y);
			node[n-i+1]=Xl(x,y);
		}
		if(work()) printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}
/*
1
11 29 3 88 16 89 33 95 59 97 70 56 31 87 97 31 44 28 21 21 60 7 52 
*/

　　

bzoj2168 半平面交模板 onleft的>=改成>就过了

//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long
#define db double
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
const int maxn=500+7;
int n,m;

char cc; ll ff;
template<typename T>void read(T& aa) {
	aa=0;cc=getchar();ff=1;
	while((cc<'0'||cc>'9')&&cc!='-') cc=getchar();
	if(cc=='-') ff=-1,cc=getchar();
	while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
	aa*=ff;
}

const db eps=1e-8;
inline int dcmp(const db &x){return fabs(x)<eps? 0 : (x>0? 1:-1);}

struct Xl{
	db x,y;
	Xl(db x=0.,db y=0.):x(x),y(y){}
	Xl operator + (const Xl& b)const{return Xl(x+b.x,y+b.y);}
	Xl operator - (const Xl& b)const{return Xl(x-b.x,y-b.y);}
	Xl operator * (const db& b)const{return Xl(b*x,b*y);}
	db angle()const{return atan2(y,x);}
}node[maxn],p[maxn];

db D_(const Xl& a,const Xl& b){return a.x*b.x+a.y*b.y;}
db X_(const Xl& a,const Xl& b){return a.x*b.y-a.y*b.x;}

struct Li{
	Xl u,v; db rad;
	Li(Xl u=Xl(0.,0.),Xl v=Xl(0.,0.)):u(u),v(v){rad=(v-u).angle();}
	bool operator < (const Li& b)const{return rad<b.rad;}
}li[maxn],zz[maxn];
int tot;

Xl lijd(const Li& a,const Li& b) {
	Xl x=a.v-a.u,y=b.v-b.u,z=a.u-b.u;
	db c=X_(z,y);
	db t=c/(c+X_(y,x+z));
	return a.u+x*t;
}

bool onleft(const Xl& a,const Li& l){return X_(l.v-l.u,a-l.u)>0;}

db HPI() {
	sort(li+1,li+tot+1);
	int s=1,t=0; zz[++t]=li[1];
	For(i,2,tot) {
		while(s<t&&!onleft(p[t-1],li[i])) t--;
		while(s<t&&!onleft(p[s],li[i])) s++;
		zz[++t]=li[i];
		if(!dcmp(X_(li[i].v-li[i].u,zz[t-1].v-zz[t-1].u))) 
			if(onleft(li[i].u,zz[--t])) zz[t]=li[i];
		if(s<t) p[t-1]=lijd(zz[t],zz[t-1]);
	}
	while(s<t&&!onleft(p[t-1],li[s])) t--;
	if(t-s<2) return 0.000;
	db rs=0; p[t]=lijd(zz[t],zz[s]);
	For(i,s+1,t-1) rs+=X_(p[i]-p[s],p[i+1]-p[s]);
	return fabs(rs)/2.0;
}

int main() {
	read(n); ll x,y;
	For(i,1,n) {
		read(m);
		For(j,1,m) {
			read(x); read(y);
			node[j]=Xl(x,y);
		}
		For(j,1,m-1) li[++tot]=Li(node[j],node[j+1]);
		li[++tot]=Li(node[m],node[1]);
	}
	printf("%.3f",HPI());
	return 0;
}

　　

bzoj4445 半平面交

简单推推式子

bzoj上精度坑人，最好开long double

//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long
#define db long double
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
const int maxn=2e5+7;
const db PI=acos(-1),eps=1e-18;
int n;

char cc; ll ff;
template<typename T>void read(T& aa) {
	aa=0;cc=getchar();ff=1;
	while((cc<'0'||cc>'9')&&cc!='-') cc=getchar();
	if(cc=='-') ff=-1,cc=getchar();
	while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
	aa*=ff;
}

db pf(const db& x){return x*x;}
inline int dcmp(const long double &x){return abs(x)<eps? 0:(x>0? 1:-1);}

struct Xl{
	db x,y;
	Xl(db x=0.,db y=0.):x(x),y(y){}
	Xl operator + (const Xl& b)const{return Xl(x+b.x,y+b.y);}
	Xl operator - (const Xl& b)const{return Xl(x-b.x,y-b.y);}
	Xl operator * (const db& b)const{return Xl(x*b,y*b);}
	db len()const{return sqrt(pf(x)+pf(y));}
	double angle() const{return atan2(y,x);}
}node[maxn],p[maxn],R;
db sr;

double D_(const Xl& a,const Xl& b){return a.x*b.x+a.y*b.y;}
double X_(const Xl& a,const Xl& b){return a.x*b.y-a.y*b.x;}

struct Li{
	Xl u,v;db rad;
	Li(Xl u=Xl(0.,0.),Xl v=Xl(0.,0.)):u(u),v(v){rad=(v-u).angle();}
    bool operator < (const Li &b) const{return rad<b.rad;}
}li[maxn],zz[maxn];
int tot;

bool onleft(const Xl& a,const Li& l) { return dcmp(X_(l.v-l.u,a-l.u))>0;}

bool get_line(Xl R,db c) {
	db a=-R.y,b=R.x,x,y;
	if(dcmp(a)==0&&dcmp(b)==0) {
		if(dcmp(c)>=0) return 1;
		return 0;
	}
	Xl u,v;
	if(dcmp(a)==0) {
		y=-c/b;
		u=Xl(0,y); v=Xl(1,y);
		if(dcmp(b)<0) swap(u,v);
	}
	else if(dcmp(b)==0) {
		x=-c/a;
		u=Xl(x,1); v=Xl(x,0);
		if(dcmp(a)<0) swap(u,v);
	}
	else {
		x=0; y=-c/b;
		u=Xl(x,y);
		x=1; y=(-a-c)/b;
		v=Xl(x,y);
		if(dcmp(b)<0) swap(u,v);
	}
	li[++tot]=Li(u,v);
	return 1;
}

Xl lijd(const Li& a,const Li& b) {
	Xl x=a.v-a.u,y=b.v-b.u,z=a.u-b.u;
	db S=X_(z,y);
	db t=S/(S+X_(y,x+z));
	return a.u+x*t;
}

db HPL() {
	sort(li+1,li+tot+1);
	int s=1,t=0;
	zz[++t]=li[1];
	For(i,2,tot) {
		while(s<t&&!onleft(p[t-1],li[i])) t--;
		while(s<t&&!onleft(p[s],li[i])) s++;
		zz[++t]=li[i];
		if(dcmp(X_(li[i].v-li[i].u,zz[t-1].v-zz[t-1].u))==0)
			if(onleft(li[i].u,zz[--t])) zz[t]=li[i];
		if(s<t) p[t-1]=lijd(zz[t],zz[t-1]);
	}
	while(s<t&&!onleft(p[t-1],li[1])) t--;
	if(s+1>t-1) return 0.0;
	db rs=0; p[t]=lijd(zz[t],zz[s]);
	For(i,s+1,t-1) rs+=X_(p[i]-p[s],p[i+1]-p[s]);
	return dcmp(rs)==0? 0:fabs(rs)/2.0;
}

db work() {
	For(i,2,n-1) {
		R=node[1]-node[2]-node[i]+node[i+1];
		sr=X_(node[i],node[i+1])-X_(node[1],node[2]);
		if(!get_line(R,sr)) return 0.0;
	}
	R=node[1]-node[2]-node[n]+node[1];
	sr=X_(node[n],node[1])-X_(node[1],node[2]);
	if(!get_line(R,sr)) return 0.0;
	return HPL();
}

int main() {
	read(n); ll x,y;
	For(i,1,n) {
		read(x); read(y);
		node[i]=Xl(x,y);
	}
	db S=0;
	For(i,2,n-1) S+=X_(node[i]-node[1],node[i+1]-node[1]);
	S=fabs(S)/2.0;
	For(i,1,n-1) li[i]=Li(node[i],node[i+1]);
	li[n]=Li(node[n],node[1]); tot=n;
	printf("%.4Lf",work()/S);
	return 0;
}
/*
3
0 0
1 0
0 1
*/

 

hdu3007 最小圆覆盖模板

数学一本通上的问题分析要把人绕晕，看代码还是要直观些

这个做法复杂度是O(n)的，当加入圆的顺序随机时，因为三点定一圆，所以不在圆内概率是3/i

所以可以用random_shuffle()这个函数让点随机(要用到algorithm这个头文件)

//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long
#define db double
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
const int maxn=500+7;
int n;

char cc; ll ff;
template<typename T>void read(T& aa) {
	aa=0;cc=getchar();ff=1;
	while((cc<'0'||cc>'9')&&cc!='-') cc=getchar();
	if(cc=='-') ff=-1,cc=getchar();
	while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
	aa*=ff;
}

const db eps=1e-10;
db pf(db x){return x*x;}
inline int dcmp(const long db &x){return abs(x)<eps? 0:(x>0? 1:-1);}

struct Xl{
	db x,y;
	Xl(db x=0.,db y=0.):x(x),y(y){}
	Xl operator + (const Xl& b)const{return Xl(x+b.x,y+b.y);}
	Xl operator - (const Xl& b)const{return Xl(x-b.x,y-b.y);}
	Xl operator * (const db& b)const{return Xl(x*b,y*b);}
	Xl operator / (const db& b)const{return Xl(x/b,y/b);}
	db len()const{return sqrt(pf(x)+pf(y));}
}node[maxn],c;

db D_(const Xl& a,const Xl& b){return a.x*b.x+a.y*b.y;}
db X_(const Xl& a,const Xl& b){return a.x*b.y-a.y*b.x;}

db r;

Xl cir_enter(const Xl& a,const Xl& b,const Xl& c) {
	db bx=b.x-a.x,by=b.y-a.y,bpf=pf(bx)+pf(by);
    db cx=c.x-a.x,cy=c.y-a.y,cpf=pf(cx)+pf(cy);
    db area=2*(bx*cy-by*cx);
    db x=(cy*bpf-by*cpf)/area;
    db y=(bx*cpf-cx*bpf)/area;
    return Xl(x+a.x,y+a.y);
}

void mincir() {
	random_shuffle(node+1,node+n+1);
	c=node[1]; r=0;
	For(i,2,n) if(dcmp((node[i]-c).len()-r)>0) {
		c=node[i]; r=0;
		For(j,1,i-1) if(dcmp((node[j]-c).len()-r)>0) {
			c=(node[i]+node[j])/2;
			r=(node[i]-c).len();
			For(k,1,j-1) if(dcmp((node[k]-c).len()-r)>0) {
				c=cir_enter(node[i],node[j],node[k]);
				r=(node[i]-c).len();
			}
		}
	}
}

int main() {
	read(n);db x,y;
	while(n) {
		For(i,1,n) {
			scanf("%lf%lf",&x,&y);
			node[i]=Xl(x,y);
		}
		mincir();
		printf("%.2f %.2f %.2f\n",c.x,c.y,r);
		read(n);
	}
	return 0;
}
/*
3
1.00 1.00
3.00 0.00
3.00 2.00
0
*/

　　

bzoj4750 妖怪

上凸包

这道题我的思维方式有点奇怪

对于每个妖怪我们看做一个点$(x,y)$

那么它的战斗力为$\frac{a+b}{a} \times x + \frac{a+b}{b} \times y$

我们考虑对于一个固定的$(a,b)$，战斗力最大的是上凸壳上的一个点

现在反过来考虑对于一个点$(x_0,y_0)$，什么时候它是战斗力最大的

令$r=\frac{b}{a}$

那么有$r \times x_0 + y_0 > r \times x_1 + y_1 $

1) $x_0 > x_1$：$r \geq - \frac{y_0-y_1}{x_0-x_1}$

2) $x_0 < x_1$：$r \leq - \frac{y_1-y_0}{x_1-x_0}$

我们发现，对于上凸壳的一个点a，假如它左边的点是b，右边的点是c

那么$r$就是在$-angle(a-b)$到$-angle(c-a)$之间，并且$r \geq 0$（因为$a,b \geq 0$）

对于所有点中一个战斗力最大的那个点$(x,y)$，什么样的$(a,b)$可以使得它的战斗力尽量小呢

$\frac{a+b}{a} \times x = \frac{a+b}{b} \times y$

$\frac{b}{a} = \frac{y}{x}$

//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long
#define db double
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
const int maxn=1e6+7;
const db INF=1e16;
int n;
db ans=INF;

char cc; ll ff;
template<typename T>void read(T& aa) {
	aa=0;cc=getchar();ff=1;
	while((cc<'0'||cc>'9')&&cc!='-') cc=getchar();
	if(cc=='-') ff=-1,cc=getchar();
	while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
	aa*=ff;
}

const db eps=1e-12;
int dcmp(const db &x){return fabs(x)<=eps? 0:(x<0? -1:1);}

struct Xl{
	db x,y;
	Xl(db x=0.,db y=0.):x(x),y(y){}
	Xl operator - (const Xl& b) const{return Xl(x-b.x,y-b.y);}
	bool operator < (const Xl& b) const{return dcmp(x-b.x)==0? y<b.y:x<b.x;}
	db angle() const{return y/x;}
}node[maxn],zz[maxn];

db X_(const Xl& a,const Xl& b) {return a.x*b.y-a.y*b.x;}

bool onleft(const Xl& p,const Xl& u,const Xl& v) {
	return dcmp(X_(v-u,p-u))>=0;
}

db p[maxn];

db get_ans(const Xl& p,db l,db r) {
	l=-l; r=-r; l=max(l,0.0); r=max(r,0.0);
	db t=p.y/p.x;
	if(t>r) t=r;
	else if(t<l) t=l;
	if(dcmp(t)==0) return INF;
	db a=100,b=a*t;
	return (a+b)/a*p.x+(a+b)/b*p.y;
}

void solve() {
	sort(node+1,node+n+1);
	int t=0;
	For(i,1,n) {
		while(t&&node[i].x==zz[t].x) t--;
		while(t>1&&onleft(node[i],zz[t-1],zz[t])) t--;
		zz[++t]=node[i];
	}
	For(i,1,t-1) p[i]=(zz[i+1]-zz[i]).angle();
	p[t]=-INF;
	For(i,1,t) ans=min(ans,get_ans(zz[i],p[i-1],p[i]));
}

int main() {
	read(n); db x,y;
	For(i,1,n) {
		read(x); read(y);
		node[i]=Xl(x,y);
	}
	solve();
	printf("%.4f",ans);
	return 0;
}

　　

bzoj1069 最大土地面积

没有什么脑子的模板题，一直以为这道题很难，然后今天看了看数据范围……

//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long
#define db double
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
const int maxn=4000+7;
int n;
db ans[maxn][maxn];

char cc; ll ff;
template<typename T>void read(T& aa) {
	aa=0;cc=getchar();ff=1;
	while((cc<'0'||cc>'9')&&cc!='-') cc=getchar();
	if(cc=='-') ff=-1,cc=getchar();
	while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
	aa*=ff;
}

const db eps=1e-10;
db pf(db x){return x*x;}
int dcmp(db x){return fabs(x)<=eps? 0:(x>0? 1:-1);}

struct Xl{
	db x,y;
	Xl(db x=0.,db y=0.):x(x),y(y){}
	bool operator < (const Xl& b) const{return dcmp(x-b.x)==0? y<b.y:x<b.x;}
	Xl operator + (const Xl& b) const{return Xl(x+b.x,y+b.y);}
	Xl operator - (const Xl& b) const{return Xl(x-b.x,y-b.y);}
	db len() const{return pf(x)+pf(y);}
}node[maxn],zz[maxn];
int t;

db D_(const Xl& a,const Xl& b) {return a.x*b.x+a.y*b.y;}
db X_(const Xl& a,const Xl& b) {return a.x*b.y-a.y*b.x;}

bool cmp(const Xl& a,const Xl& b) {
	return dcmp(X_(a-node[1],b-node[1]))==0?
			(a-node[1]).len()<(b-node[1]).len() : dcmp(X_(a-node[1],b-node[1]))>0;
}

void graham() {
	For(i,2,n) if(node[i]<node[1]) swap(node[i],node[1]);
	sort(node+2,node+n+1,cmp);
	zz[++t]=node[1]; zz[++t]=node[2];
	For(i,3,n) {
		while(t>1&&dcmp(X_(zz[t]-zz[t-1],node[i]-zz[t])<=0)) --t;
		zz[++t]=node[i];
	}
}

void RC() {
	For(i,1,t) zz[i+t]=zz[i]; zz[0]=zz[t];
	int now; Xl p;
	For(i,1,t) {
		now=(i+1)%t;
		For(j,i+1,i+t-1) {
			p=zz[j]-zz[i];
			while(X_(p,zz[now]-zz[i])<X_(p,zz[now+1]-zz[i])) now=(now+1)%t;
			ans[i][j]=X_(p,zz[now]-zz[i]);
		}
	} 
}

int main() {
	read(n); db x,y;
	For(i,1,n) {
		scanf("%lf%lf",&x,&y);
		node[i]=Xl(x,y);
	}
	graham();
	RC(); db totans=0;
	For(i,1,t) For(j,i+1,t) 
		totans=max(totans,ans[i][j]+ans[j][i+t]);
	printf("%.3lf",totans/2);
	return 0;
}

　　

bzoj2388 旅行规划

分块+凸包+神奇的三分

//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long
#define db double
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
const int maxn=2e5+7,maxt=500+7;
const ll INF=1e16;
int n,m,sz,tot;
ll s[maxn],A[maxn],B[maxn];

char cc; ll ff;
template<typename T>void read(T& aa) {
	aa=0;cc=getchar();ff=1;
	while((cc<'0'||cc>'9')&&cc!='-') cc=getchar();
	if(cc=='-') ff=-1,cc=getchar();
	while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
	aa*=ff;
}

const db eps=1e-8;
int dcmp(db x) {return fabs(x)<=eps? 0:(x<0? -1:1);}

struct Xl{
	db x,y;
	Xl(db x=0.,db y=0.):x(x),y(y){}
	Xl operator - (const Xl& b) const{return Xl(x-b.x,y-b.y);}
	bool operator < (const Xl& b) const{return dcmp(x-b.x==0)? y<b.y : x<b.x;}
}node[maxn],p[maxt][maxt];
int top[maxt];

db X_(const Xl& a,const Xl& b) {return a.x*b.y-a.y*b.x;}
db D_(const Xl& a,const Xl& b) {return a.x*b.x+a.y*b.y;}

void get_ham(int pos) {
	int l=max(pos*sz,1),r=min((pos+1)*sz-1,n);
	For(i,l,r) node[i]=Xl(i,s[i]);
	int t=0; p[pos][++t]=node[l];
	For(i,l+1,r) {
		while(t>1&&dcmp(X_(p[pos][t]-p[pos][t-1],node[i]-p[pos][t-1]))>=0) t--;
		p[pos][++t]=node[i];
	}
	top[pos]=t;
}

void pd(int pos) {
	if(A[pos]==0&&B[pos]==0) return;
	int l=max(pos*sz,1),r=min((pos+1)*sz-1,n);
	For(i,l,r) s[i]+=A[pos]+(i-l)*B[pos];
	A[pos]=B[pos]=0;
}

void chge(int l,int r,ll x) {
	pd(l/sz); pd(r/sz);
	if(l/sz==r/sz) {
		For(i,l,r) s[i]+=x*(i-l+1);
		get_ham(l/sz);
		For(i,r+1,min((r/sz+1)*sz-1,n)) s[i]+=x*(r-l+1);
		For(i,l/sz+1,tot) A[i]+=(r-l+1)*x;
		return;
	}
	For(i,l,min((l/sz+1)*sz-1,n)) s[i]+=x*(i-l+1);
	get_ham(l/sz);
	For(i,max((r/sz)*sz,1),r) s[i]+=x*(i-l+1);
	For(i,r+1,min((r/sz+1)*sz-1,n)) s[i]+=x*(r-l+1);
	get_ham(r/sz);
	For(i,l/sz+1,r/sz-1) A[i]+=(i*sz-l+1)*x,B[i]+=x;
	For(i,r/sz+1,tot) A[i]+=(r-l+1)*x;
}

ll cal(int pos) {return A[pos/sz]+B[pos/sz]*(pos-(pos/sz)*sz)+s[pos];}

ll get_ans(int pos) {
	int l=1,r=top[pos],mid; ll a,b,c;
	while(l<=r) {
		if(l+1>=r) return max(cal(p[pos][l].x),cal(p[pos][r].x));
		mid=(l+r)>>1;
		a=cal(p[pos][mid-1].x);
		b=cal(p[pos][mid].x);
		c=cal(p[pos][mid+1].x);
		if(a<b&&b<c) l=mid+1;
		else if(a>b&&b>c) r=mid-1;
		else return b;
	}
	return 0;
}

ll Yth(int l,int r) {
	pd(l/sz); pd(r/sz);
	ll rs=-INF;
	if(l/sz==r/sz) {
		For(i,l,r) rs=max(rs,s[i]);
		return rs;
	}
	For(i,l,min((l/sz+1)*sz-1,n)) rs=max(rs,s[i]);
	For(i,max((r/sz)*sz,1),r) rs=max(rs,s[i]);
	For(i,l/sz+1,r/sz-1) rs=max(rs,get_ans(i));
	return rs;
}

int main() {
	read(n); sz=min((int)sqrt(n),200)+1; tot=n/sz;
	ll op,l,r,x;
	For(i,1,n) read(s[i]),s[i]+=s[i-1];
	For(i,0,tot) get_ham(i);
	read(m); 
	For(i,1,m) {
		read(op); read(l); read(r);
		if(op) printf("%lld\n",Yth(l,r));
		else {
			read(x);
			chge(l,r,x);
		}
	}
	return 0;
}

 

bzoj1038瞭望塔

半平面交+线段最小距离 板子

//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long
#define db double
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
const int maxn=1000+7;
int n;

char cc; ll ff;
template<typename T>void read(T& aa) {
	aa=0;cc=getchar();ff=1;
	while((cc<'0'||cc>'9')&&cc!='-') cc=getchar();
	if(cc=='-') ff=-1,cc=getchar();
	while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
	aa*=ff;
}

const db eps=1e-14,INF=1e30;
int dcmp(const db x){return fabs(x)<=eps? 0:(x>0? 1:-1);}
db pf(const db x){return x*x;}

struct Xl{
	db x,y;
	Xl(db x=0.,db y=0.):x(x),y(y){}
	Xl operator + (const Xl& b) const{return Xl(x+b.x,y+b.y);}
	Xl operator - (const Xl& b) const{return Xl(x-b.x,y-b.y);}
	Xl operator * (const db& b) const{return Xl(x*b,y*b);}
	Xl operator / (const db& b) const{return Xl(x/b,y/b);}
	db angle() const{return atan2(y,x);}
	db len() const{return sqrt(pf(x)+pf(y));}
}node[maxn],p[maxn];

db X_(const Xl& a,const Xl& b) {return a.x*b.y-a.y*b.x;}
db D_(const Xl& a,const Xl& b) {return a.x*b.x+a.y*b.y;}

struct Li{
	Xl u,v; db rad;
	Li(){}
	Li(Xl u,Xl v):u(u),v(v){rad=(v-u).angle();}
	bool operator < (const Li& b) const{return rad<b.rad;}
}li[maxn],zz[maxn];

bool onleft(const Xl& a,const Li& l) {return dcmp(X_(l.v-l.u,a-l.u))>0;}
Xl lijd(const Li& a,const Li& b) {
	Xl x=a.v-a.u,y=b.v-b.u,z=a.u-b.u;
	db t=X_(z,y);
	t=t/(t+X_(y,x+z));
	return a.u+x*t;
}

db xdjl(Li a,Li b) {
	return min((a.u-b.u).len(),(a.v-b.v).len());
}

db get_ans(db l,db r,Li a,Li b) {
	if(l>r) return INF;
	Xl x=a.v-a.u,y=b.v-b.u;
	a.u=a.u+x*((l-a.u.x)/(a.v.x-a.u.x));
	b.u=b.u+y*((l-b.u.x)/(b.v.x-b.u.x));
	x=a.v-a.u; y=b.v-b.u;
	a.v=a.u+x*((r-a.u.x)/(a.v.x-a.u.x));
	b.v=b.u+y*((r-b.u.x)/(b.v.x-b.u.x));
	return xdjl(a,b);
}

db HPI() {
	sort(li+1,li+n);
	int s=1,t=0; zz[++t]=li[1];
	For(i,2,n-1) {
		while(s<t&&!onleft(p[t-1],li[i])) t--;
		while(s<t&&!onleft(p[s],li[i])) s++;
		zz[++t]=li[i];
		if(s<t&&dcmp(li[i].rad-zz[t-1].rad)==0) if(onleft(li[i].u,zz[--t])) zz[t]=li[i];
		if(s<t) p[t-1]=lijd(zz[t],zz[t-1]);
	}
	while(s<t&&!onleft(p[t-1],li[s])) t--;
	p[s-1].x=-INF;
	p[t+1].x=INF;
	db rs=INF,l,r; int pos=s;
	For(i,1,n-1) {
		while(pos<t&&p[pos].x<node[i].x) ++pos;
		l=max(node[i].x,p[pos-1].x); r=min(node[i+1].x,p[pos].x);
		rs=min(rs,get_ans(l,r,zz[pos],Li(node[i],node[i+1])));
		while(pos<t&&p[pos].x<=node[i+1].x) {
			++pos;
			l=max(node[i].x,p[pos-1].x); r=min(node[i+1].x,p[pos].x);
			rs=min(rs,get_ans(l,r,zz[pos],Li(node[i],node[i+1])));
		}
	}
	return rs;
}

int main() {
	read(n);
	For(i,1,n) scanf("%lf",&node[i].x);
	For(i,1,n) scanf("%lf",&node[i].y);
	For(i,1,n-1) li[i]=Li(node[i],node[i+1]);
	printf("%.3f",HPI()); 
	return 0;
}

　　

转载于:https://www.cnblogs.com/Serene-shixinyi/p/8462437.html

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