题目大意：给定一个平面，多次插入一条线段，以及询问某个x值能截到的最大纵坐标

OTZ

一份详细的网址：http://hi.baidu.com/wyl8899/item/2deafd3a376ef2d46d15e998

注意细节

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define M 100100
#define EPS 1e-7
using namespace std;
struct Line{
	double k,b;
	Line() {}
	Line(double _,double __):
		k(_),b(__) {}
	double F(int x)
	{
		return k*x+b;
	}
	friend double Get_Intersection(const Line &l1,const Line &l2)
	{
		return -(l1.b-l2.b)/(l1.k-l2.k);
	}
}lines[M];int tot;

int Max(int x,int y,int pos)
{
	double f1=lines[x].F(pos);
	double f2=lines[y].F(pos);
	if(fabs(f1-f2)>EPS)
		return f1>f2?x:y;
	return min(x,y);
}

struct Segtree{
	Segtree *ls,*rs;
	int line;
	Segtree(Segtree *_,Segtree *__,int ___):
		ls(_),rs(__),line(___) {}
	void Insert(int x,int y,int l,int r,int val)
	{
		int mid=x+y>>1;
		if(x==l&&y==r)
		{
			if( lines[line].F(x)>lines[val].F(x)+EPS && lines[line].F(y)>lines[val].F(y)+EPS )
				return ;
			if( lines[line].F(x)<lines[val].F(x)-EPS && lines[line].F(y)<lines[val].F(y)-EPS )
			{
				line=val;
				return ;
			}
			if( fabs(lines[line].F(x)-lines[val].F(x))<EPS && fabs(lines[line].F(y)-lines[val].F(y))<EPS )
			{
				line=min(line,val);
				return ;
			}
			double temp=Get_Intersection(lines[line],lines[val]);
			if(temp<=mid+0.5)
			{
				if( lines[line].F(x) > lines[val].F(x)-EPS )
					ls->Insert(x,mid,l,mid,line),line=val;
				else
					ls->Insert(x,mid,l,mid,val);
			}
			else
			{
				if( lines[line].F(y) > lines[val].F(y)-EPS )
					rs->Insert(mid+1,y,mid+1,r,line),line=val;
				else
					rs->Insert(mid+1,y,mid+1,r,val);
			}
			return ;
		}
		if(r<=mid) ls->Insert(x,mid,l,r,val);
		else if(l>mid) rs->Insert(mid+1,y,l,r,val);
		else ls->Insert(x,mid,l,mid,val) , rs->Insert(mid+1,y,mid+1,r,val);
	}
	int Query(int x,int y,int pos)
	{
		int mid=x+y>>1;
		if(x==y) return line;
		if(pos<=mid) return Max(ls->Query(x,mid,pos),line,pos);
		else return Max(rs->Query(mid+1,y,pos),line,pos);
	}
}*tree;
Segtree* Build_Tree(int x,int y)
{
	int mid=x+y>>1;
	if(x==y) return new Segtree(0x0,0x0,0);
	return new Segtree(Build_Tree(x,mid),Build_Tree(mid+1,y),0);
}

int n,last_ans;

int main()
{
	int i,p,x,x1,y1,x2,y2;
	cin>>n;
	tree=Build_Tree(1,39989);
	for(i=1;i<=n;i++)
	{
		scanf("%d",&p);
		if(p==0)
		{
			scanf("%d",&x);
			x=x=((x+last_ans-1)%39989+1);
			printf("%d\n",last_ans=tree->Query(1,39989,x) );
		}
		else
		{
			scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
			x1=(x1+last_ans-1)%39989+1;
			y1=(y1+last_ans-1)%1000000000+1;
			x2=(x2+last_ans-1)%39989+1;
			y2=(y2+last_ans-1)%1000000000+1;
			if(x1>x2) swap(x1,x2),swap(y1,y2);
			if(x1==x2)
				new(&lines[++tot])Line(0,max(y1,y2));
			else
			{
				double k=(double)(y1-y2)/(x1-x2);
				new(&lines[++tot])Line(k,y1-k*x1);
			}
			tree->Insert(1,39989,x1,x2,tot);
		}
	}
	return 0;
}

BZOJ 3165 Heoi2013 Segment 线段树