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  • bzoj[2402] 陶陶的难题II 树链剖分+线段树+二分答案+凸包

    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 }
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  • 原文地址:https://www.cnblogs.com/lnxcj/p/10158418.html
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