[点分治 离线] BZOJ 4449 [Neerc2015]Distance on Triangulation

其实简单的找一条边然后分成两半
被我硬生生写成对偶图点分治分成三块 233
就把所有询问一起处理
点分治时 如果两个点在同一侧 就递归 否则对三个点bfs就能知道最短路径

#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>#include<map>#include<vector>#define cl(x) memset(x,0,sizeof(x))using namespace std;typedef pair<int,int> abcd;inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void read(int &x){  char c=nc(),b=1;  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;  for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}const int N=100005;struct edge{  int u,v,next;}G[N<<3];int head[N],head2[N],inum;inline void add(int u,int v,int *head){  int p=++inum; G[p].u=u; G[p].v=v; G[p].next=head[u]; head[u]=p;}inline void link(int u,int v,int *head){  add(u,v,head); add(v,u,head);}#define V G[p].vint n;int deg[N];int pt[N][3];map<abcd,int> ed;inline void BT(){  static int Q[N],l,r,rem[N];  l=r=-1; cl(rem);  for (int i=1;i<=n;i++) if (deg[i]==2) Q[++r]=i;  for (int i=1;i<=n-2;i++){    int u=Q[++l],a[3]; rem[u]=1;    *a=0;    for (int p=head[u];p;p=G[p].next)      if (!rem[V])    a[++*a]=V;    pt[i][0]=u; pt[i][1]=a[1]; pt[i][2]=a[2];    if ((--deg[a[1]])==2) Q[++r]=a[1];    if ((--deg[a[2]])==2) Q[++r]=a[2];  }  for (int s=1;s<=n-2;s++){    for (int i=0;i<3;i++)      for (int j=0;j<i;j++){    int u=pt[s][i],v=pt[s][j];    if (u>v) swap(u,v);    if (ed.count(abcd(u,v)))      link(s,ed[abcd(u,v)],head2);    else      ed[abcd(u,v)]=s;      }  }}int del[N];int sum,minv,rt;int size[N];inline void Root(int u,int fa){  size[u]=1; int maxv=0;  for (int p=head2[u];p;p=G[p].next)    if (V!=fa && !del[V])      Root(V,u),size[u]+=size[V],maxv=max(maxv,size[V]);  maxv=max(maxv,sum-size[u]);  if (maxv<minv) minv=maxv,rt=u;}int c[N],col=4;int ins[N],clk;int dis[3][N];inline void bfs(int S,int *dis){  static int Q[N],l,r;  l=r=-1; ++clk;  Q[++r]=S; dis[S]=0; ins[S]=clk;  while (l<r){    int u=Q[++l];    for (int p=head[u];p;p=G[p].next)      if (c[V]==c[u] && ins[V]!=clk){    dis[V]=dis[u]+1;    ins[V]=clk;    Q[++r]=V;      }  }}int Col;inline void color(int u,int fa){  size[u]=1;  for (int i=0;i<3;i++) c[pt[u][i]]=Col;  for (int p=head2[u];p;p=G[p].next)    if (V!=fa && !del[V])      color(V,u),size[u]+=size[V];}struct event{  int u,v,idx;  event(int u=0,int v=0,int idx=0):u(u),v(v),idx(idx){ }};int Ans[N<<1];inline void Divide(int u,vector<event> &que){  if (que.empty()) return;  del[u]=1;  for (int i=0;i<3;i++)    bfs(pt[u][i],dis[i]);  Col=0;  for (int p=head2[u];p;p=G[p].next)    if (!del[V]){      Col++;      color(V,0);    }  for (int i=0;i<3;i++) c[pt[u][i]]=0;  vector<event> vec[4];  for (int i=0;i<(int)que.size();i++){    int a=que[i].u,b=que[i].v,idx=que[i].idx;    if (c[a]==0){      for (int j=0;j<3;j++)    if (a==pt[u][j])      Ans[idx]=dis[j][b];    }else if (c[b]==0){      for (int j=0;j<3;j++)    if (b==pt[u][j])      Ans[idx]=dis[j][a];    }else if (c[a]!=c[b]){      Ans[idx]=1<<30;      for (int j=0;j<3;j++)    Ans[idx]=min(Ans[idx],dis[j][a]+dis[j][b]);    }else{      vec[c[a]].push_back(que[i]);    }  }  for (int p=head2[u],i=0;p;p=G[p].next)    if (!del[V]){      Col=++col; color(V,0);      sum=size[V]; minv=1<<30;      Root(V,0);      Divide(rt,vec[++i]);    }}int main(){  int iu,iv,Q;  freopen("drive.in","r",stdin);  freopen("drive.out","w",stdout);  read(n);  for (int i=1;i<n;i++) link(i,i+1,head);  link(n,1,head);  for (int i=1;i<=n;i++) deg[i]=2;   for (int i=1;i<=n-3;i++)    read(iu),read(iv),link(iu,iv,head),deg[iu]++,deg[iv]++;  BT();  read(Q);  vector<event> vec;  for (int i=1;i<=Q;i++)    read(iu),read(iv),vec.push_back(event(iu,iv,i));  sum=n-2; minv=1<<30;  Root(1,0);  for (int i=1;i<=n;i++) c[i]=col;  Divide(rt,vec);  for (int i=1;i<=Q;i++)    printf("%d\n",Ans[i]);  return 0;}