bzoj[2402] 陶陶的难题II 树链剖分+线段树+二分答案+凸包

Posted 2021-01-28 lnxcj

将y+q/x+p的值设为x

由于i,j互不干扰,所以我们可以将x,y p,q拉出来分别计算

问题转化为存在x,y,满足y-mid*x+q-mid*p>=0的情况下mid最大

不难发现答案具有单调性,于是二分答案

显然要取出一对(x,y)或(p,q)使y(q)-x(p)*mid尽可能大

以x,y举例,若取点1比取点2更优(x1>x2),则(y1-y2)/(x1-x2)>=mid

用单调栈维护上凸包,存到线段树中

查询时二分凸包

写麻烦了。。。

  1 #include<cstdio>
  2 #include<vector>
  3 #include<cstring>
  4 #include<algorithm>
  5 using namespace std;
  6 int n,m,cnt,tot,tp;
  7 double ans;
  8 struct node{
  9     double x,y;
 10 }pox[30005],poq[30005],mst;
 11 vector<node>segx[30000*5+5];
 12 vector<node>segq[30000*5+5];
 13 int ch[30005];
 14 node tm1[30005],tm2[30005];
 15 int pos[30005],grand[30005];
 16 int head[30005],dep[30005];
 17 int siz[30005],son[30005];
 18 int fa[30005],stk[30005];
 19 struct Edge{
 20     int fr;
 21     int to;
 22     int nxt;
 23 }edge[60005];
 24 bool operator == (node a,node b){
 25     return a.x-b.x<=0.0000001&&b.x-a.x<=0.0000001&&a.y-b.y<=0.0000001&&b.y-a.y<=0.0000001;
 26 }
 27 int cmp(node a,node b){
 28     return a.x<b.x;
 29 }
 30 void init(){
 31     dep[1]=1;
 32     memset(head,-1,sizeof(head));
 33 }
 34 void addedge(int u,int v){
 35     edge[cnt].fr=u;
 36     edge[cnt].to=v;
 37     edge[cnt].nxt=head[u];
 38     head[u]=cnt++;
 39 }
 40 void dfs1(int u){
 41     siz[u]=1;
 42     for(int i=head[u];i!=-1;i=edge[i].nxt){
 43         int v=edge[i].to;
 44         if(v==fa[u])continue;
 45         fa[v]=u;dep[v]=dep[u]+1;
 46         dfs1(v);siz[u]+=siz[v];
 47         if(siz[v]>siz[son[u]])son[u]=v;
 48     }
 49 }
 50 void dfs2(int u){
 51     if(son[fa[u]]!=u)grand[u]=u;
 52     else grand[u]=grand[fa[u]];
 53     stk[++tp]=u;pos[u]=tp;
 54     if(son[u])dfs2(son[u]);
 55     for(int i=head[u];i!=-1;i=edge[i].nxt){
 56         int v=edge[i].to;
 57         if(v==fa[u])continue;
 58         if(v!=son[u])dfs2(v);
 59     }
 60 }
 61 int lca(int x,int y){
 62     while(grand[x]!=grand[y]){
 63         if(dep[grand[x]]<dep[grand[y]])swap(x,y);
 64         x=fa[grand[x]];
 65     }
 66     if(dep[x]>dep[y])swap(x,y);
 67     return x;
 68 }
 69 void build(int u,int l,int r){
 70     for(int i=l;i<=r;i++){
 71         tm1[i]=pox[stk[i]];
 72         tm2[i]=poq[stk[i]];
 73     }
 74     sort(tm1+l,tm1+r+1,cmp);
 75     sort(tm2+l,tm2+r+1,cmp);
 76     tp=0;
 77     for(int i=l;i<=r;i++){
 78         while(tp>1&&1ll*(tm1[ch[tp]].y-tm1[ch[tp-1]].y)*(tm1[i].x-tm1[ch[tp]].x)<1ll*(tm1[i].y-tm1[ch[tp]].y)*(tm1[ch[tp]].x-tm1[ch[tp-1]].x)){
 79             tp--;
 80         }ch[++tp]=i;
 81     }
 82     for(int i=1;i<=tp;i++)segx[u].push_back(tm1[ch[i]]);
 83     tp=0;
 84     for(int i=l;i<=r;i++){
 85         while(tp>1&&1ll*(tm2[ch[tp]].y-tm2[ch[tp-1]].y)*(tm2[i].x-tm2[ch[tp]].x)<1ll*(tm2[i].y-tm2[ch[tp]].y)*(tm2[ch[tp]].x-tm2[ch[tp-1]].x)){
 86             tp--;
 87         }ch[++tp]=i;
 88     }
 89     for(int i=1;i<=tp;i++)segq[u].push_back(tm2[ch[i]]);
 90     if(l==r)return;
 91     int mid=(l+r)>>1;
 92     build(u<<1,l,mid);
 93     build(u<<1|1,mid+1,r);
 94 }
 95 node queryx(int u,int l,int r,int ql,int qr,double qv){
 96     if(l>=ql&&r<=qr){
 97         int le=0,ri=segx[u].size()-1;
 98         while(le<=ri){
 99             int mid=(le+ri)>>1;
100             if(!mid)le=mid+1;
101             else if(1ll*(segx[u][mid].y-segx[u][mid-1].y)>1ll*(segx[u][mid].x-segx[u][mid-1].x)*qv)le=mid+1;
102             else ri=mid-1;
103         }
104         return segx[u][ri];
105     }int mid=(l+r)>>1;
106     if(qr<=mid)return queryx(u<<1,l,mid,ql,qr,qv);
107     else if(ql>mid)return queryx(u<<1|1,mid+1,r,ql,qr,qv);
108     else{
109         node a=queryx(u<<1,l,mid,ql,qr,qv);
110         node b=queryx(u<<1|1,mid+1,r,ql,qr,qv);
111         if(a.x<b.x)swap(a,b);
112         node ret=(1ll*(a.y-b.y)<1ll*qv*(a.x-b.x))?b:a;
113         return ret;
114     }
115 }
116 node queryq(int u,int l,int r,int ql,int qr,double qv){
117     if(l>=ql&&r<=qr){
118         int le=0,ri=segq[u].size()-1;
119         while(le<=ri){
120             int mid=(le+ri)>>1;
121             if(!mid)le=mid+1;
122             else if(1ll*(segq[u][mid].y-segq[u][mid-1].y)>1ll*(segq[u][mid].x-segq[u][mid-1].x)*qv)le=mid+1;
123             else ri=mid-1;
124         }
125         return segq[u][ri];
126     }int mid=(l+r)>>1;
127     if(qr<=mid)return queryq(u<<1,l,mid,ql,qr,qv);
128     else if(ql>mid)return queryq(u<<1|1,mid+1,r,ql,qr,qv);
129     else{
130         node a=queryq(u<<1,l,mid,ql,qr,qv);
131         node b=queryq(u<<1|1,mid+1,r,ql,qr,qv);
132         if(a.x<b.x)swap(a,b);
133         node ret=(1ll*(a.y-b.y)<1ll*qv*(a.x-b.x))?b:a;
134         return ret;
135     }
136 }
137 int main(){
138     init();
139     scanf("%d",&n);
140     for(int i=1;i<=n;i++)scanf("%lf",&pox[i].x);
141     for(int i=1;i<=n;i++)scanf("%lf",&pox[i].y);
142     for(int i=1;i<=n;i++)scanf("%lf",&poq[i].x);
143     for(int i=1;i<=n;i++)scanf("%lf",&poq[i].y);
144     for(int i=1;i<n;i++){
145         int u,v;
146         scanf("%d%d",&u,&v);
147         addedge(u,v);
148         addedge(v,u);
149     }dfs1(1);dfs2(1);
150     build(1,1,n);
151     scanf("%d",&m);
152     for(int i=1;i<=m;i++){
153         double ans=0;
154         node ans1,ans2,tmp1,tmp2;
155         ans1=ans2=tmp1=tmp2=mst;
156         int u,v,now,f;
157         double l,r;
158         scanf("%d%d",&u,&v);
159         l=0,r=100000;f=lca(u,v);
160         while(r-l>=0.0001){
161             double mid=(l+r)/2;now=u;
162             while(dep[grand[now]]>dep[f]){
163                 tmp1=queryx(1,1,n,pos[grand[now]],pos[now],mid);
164                 tmp2=queryq(1,1,n,pos[grand[now]],pos[now],mid);
165                 if(ans1==mst)ans1=tmp1;
166                 if(ans2==mst)ans2=tmp2;
167                 if(tmp1.x>ans1.x)swap(ans1,tmp1);
168                 if(tmp2.x>ans2.x)swap(ans2,tmp2);
169                 ans1=(1ll*(ans1.y-tmp1.y)<1ll*(ans1.x-tmp1.x)*mid)?tmp1:ans1;
170                 ans2=(1ll*(ans2.y-tmp2.y)<1ll*(ans2.x-tmp2.x)*mid)?tmp2:ans2;
171                 now=fa[grand[now]];
172             }
173             tmp1=queryx(1,1,n,pos[f],pos[now],mid);
174             tmp2=queryq(1,1,n,pos[f],pos[now],mid);
175             if(ans1==mst)ans1=tmp1;
176             if(ans2==mst)ans2=tmp2;
177             if(tmp1.x>ans1.x)swap(ans1,tmp1);
178             if(tmp2.x>ans2.x)swap(ans2,tmp2);
179             ans1=(1ll*(ans1.y-tmp1.y)<1ll*(ans1.x-tmp1.x)*mid)?tmp1:ans1;
180             ans2=(1ll*(ans2.y-tmp2.y)<1ll*(ans2.x-tmp2.x)*mid)?tmp2:ans2;
181             now=v;
182             while(dep[grand[now]]>dep[f]){
183                 tmp1=queryx(1,1,n,pos[grand[now]],pos[now],mid);
184                 tmp2=queryq(1,1,n,pos[grand[now]],pos[now],mid);
185                 if(ans1==mst)ans1=tmp1;
186                 if(ans2==mst)ans2=tmp2;
187                 if(tmp1.x>ans1.x)swap(ans1,tmp1);
188                 if(tmp2.x>ans2.x)swap(ans2,tmp2);
189                 ans1=(1ll*(ans1.y-tmp1.y)<1ll*(ans1.x-tmp1.x)*mid)?tmp1:ans1;
190                 ans2=(1ll*(ans2.y-tmp2.y)<1ll*(ans2.x-tmp2.x)*mid)?tmp2:ans2;
191                 now=fa[grand[now]];
192             }
193             tmp1=queryx(1,1,n,pos[f],pos[now],mid);
194             tmp2=queryq(1,1,n,pos[f],pos[now],mid);
195             if(ans1==mst)ans1=tmp1;
196             if(ans2==mst)ans2=tmp2;
197             if(tmp1.x>ans1.x)swap(ans1,tmp1);
198             if(tmp2.x>ans2.x)swap(ans2,tmp2);
199             ans1=(1ll*(ans1.y-tmp1.y)<1ll*(ans1.x-tmp1.x)*mid)?tmp1:ans1;
200             ans2=(1ll*(ans2.y-tmp2.y)<1ll*(ans2.x-tmp2.x)*mid)?tmp2:ans2;
201             if(ans1.y-ans1.x*mid+ans2.y-ans2.x*mid>0)l=mid;
202             else r=mid;
203         }
204         printf("%.4lf
",r);
205     }
206     return 0;
207 }