BZOJ 2402 陶陶的难题II 二分答案+斜率优化+树链剖分+线段树维护凸包

题目大意：给定一棵树，每个点有两个坐标(x1,y1)和(x2,y2)，多次询问某条链上选择两个点i和j(可以相同)，求(y1i+y2j)/(x1i+x2j)的最大值

我竟没看出来这是01分数规划。。。真是老了。。。

二分答案ans，问题转化成验证(y1i+y2j)/(x1i+x2j)是否>=ans

将式子变形可得(y1i-ans*x1i)+(y2j-ans*x2j)>=0

加号两边独立，分别计算即可

问题转化为求链上y-ans*x最大的点

令P=y-ans*x 则y=ans*x+P

我们发现这是一个斜率优化的形式 因此我们使用树链剖分 用线段树维护凸包即可

每次查询 树链剖分一个log 线段树一个log 每次凸包上二分一个log 加上最外层的二分一共4个log

能过真是奇迹。。。。

#include <cmath>#include <vector>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define M 30300#define EPS 1e-7#define INF 1e10using namespace std;struct Point{double x,y;Point() {}Point(double _,double __):x(_),y(__) {}friend Point operator + (const Point &p1,const Point &p2){return Point(p1.x+p2.x,p1.y+p2.y);}friend Point operator - (const Point &p1,const Point &p2){return Point(p1.x-p2.x,p1.y-p2.y);}friend double operator * (const Point &p1,const Point &p2){return p1.x*p2.y-p1.y*p2.x;}friend bool operator < (const Point &p1,const Point &p2){return p1.x<p2.x || p1.x==p2.x&&p1.y<p2.y ;}friend double Get_Slope(const Point &p1,const Point &p2){if( fabs(p1.x-p2.x)<EPS )return p1.y<p2.y?INF:-INF;return (p1.y-p2.y)/(p1.x-p2.x);}}points1[M],points2[M];struct abcd{int to,next;}table[M<<1];int head[M],tot;int n,m;int fa[M],son[M],dpt[M],size[M];int top[M],pos[M],a[M],cnt;void Add(int x,int y){table[++tot].to=y;table[tot].next=head[x];head[x]=tot;}void DFS1(int x){int i;dpt[x]=dpt[fa[x]]+1;size[x]=1;for(i=head[x];i;i=table[i].next){if(table[i].to==fa[x])continue;fa[table[i].to]=x;DFS1(table[i].to);size[x]+=size[table[i].to];if(size[table[i].to]>size[son[x]])son[x]=table[i].to;}}void DFS2(int x){int i;a[pos[x]=++cnt]=x;if(son[fa[x]]==x)top[x]=top[fa[x]];elsetop[x]=x;if(son[x]) DFS2(son[x]);for(i=head[x];i;i=table[i].next){if(table[i].to==fa[x]||table[i].to==son[x])continue;DFS2(table[i].to);}}void Merge(const vector<Point> &h1,const vector<Point> &h2,vector<Point> &h){vector<Point>::const_iterator i,j;i=h1.begin();j=h2.begin();int top=0;while( i!=h1.end() || j!=h2.end() ){Point p=i==h1.end()?*j++:j==h2.end()?*i++:*i<*j?*i++:*j++;while( top>=2 && (h[top-1]-h[top-2])*(p-h[top-1])>-EPS )h.pop_back(),top--;h.push_back(p),top++;}}double Bisection(const vector<Point> &h,double k){int l=0,r=h.size();while(l+1<r){int mid=l+r>>1;if( Get_Slope(h[mid-1],h[mid])>k-EPS )l=mid;elser=mid;}return h[l].y-k*h[l].x;}struct Segtree{Segtree *ls,*rs;vector<Point> h1,h2;void* operator new (size_t){static Segtree mempool[M<<1],*C=mempool;return C++;}void Build_Tree(int x,int y){int mid=x+y>>1;if(x==y){h1.push_back(points1[a[mid]]);h2.push_back(points2[a[mid]]);return ;}(ls=new Segtree)->Build_Tree(x,mid);(rs=new Segtree)->Build_Tree(mid+1,y);Merge(ls->h1,rs->h1,h1);Merge(ls->h2,rs->h2,h2);}double Query(int x,int y,int l,int r,double k,bool flag){int mid=x+y>>1;if(x==l&&y==r)return !flag?Bisection(h1,k):Bisection(h2,k);if(r<=mid)return ls->Query(x,mid,l,r,k,flag);if(l>mid)return rs->Query(mid+1,y,l,r,k,flag);return max( ls->Query(x,mid,l,mid,k,flag) , rs->Query(mid+1,y,mid+1,r,k,flag) );}}*tree=new Segtree[M];double Query(int x,int y,double k,bool flag){int fx=top[x],fy=top[y];double re=-INF;while(fx!=fy){if(dpt[fx]<dpt[fy])swap(x,y),swap(fx,fy);re=max(re,tree->Query(1,n,pos[fx],pos[x],k,flag));x=fa[fx];fx=top[x];}if(dpt[x]<dpt[y])swap(x,y);re=max(re,tree->Query(1,n,pos[y],pos[x],k,flag));return re;}double Bisection(int x,int y){double l=0,r=1e8;while(r-l>1e-5){double mid=(l+r)/2.0;if( Query(x,y,mid,false) + Query(x,y,mid,true) > 0 )l=mid;elser=mid;}return (l+r)/2;}int main(){int i,x,y;cin>>n;for(i=1;i<=n;i++)scanf("%lf",&points1[i].x);for(i=1;i<=n;i++)scanf("%lf",&points1[i].y);for(i=1;i<=n;i++)scanf("%lf",&points2[i].x);for(i=1;i<=n;i++)scanf("%lf",&points2[i].y);for(i=1;i<n;i++){scanf("%d%d",&x,&y);Add(x,y);Add(y,x);}DFS1(1);DFS2(1);tree->Build_Tree(1,n);cin>>m;for(i=1;i<=m;i++){scanf("%d%d",&x,&y);printf("%.5lf\n",Bisection(x,y));}}