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\n",r);
205 }
206 return 0;
207 }
原文地址：https://www.cnblogs.com/lnxcj/p/10158418.html
时间： 2024-11-09 00:05:14