[POI2012] RAN-Rendezvous
题目描述
Byteasar is a ranger who works in the Arrow Cave - a famous rendezvous destination among lovers.
The cave consists of \(n\) chambers connected with one-way corridors.
In each chamber exactly one outgoing corridor is marked with an arrow.
Every corridor leads directly to some (not necessarily different) chamber.
The enamoured couples that agree to meet in the Arrow Cave are notorious for forgetting to agree upon specific chamber, and consequently often cannot find their dates.
In the past this led to many mix-ups and misunderstandings\dots But ever since each chamber is equipped with an emergency telephone line to the ranger on duty, helping the enamoured find their dates has become the rangers' main occupation.
The rangers came up with the following method.
Knowing where the enamoured are, they tell each of them how many times they should follow the corridor marked with an arrow in order to meet their date.
The lovers obviously want to meet as soon as possible - after all, they came to the cave to spend time together, not to wander around alone!
Most rangers are happy to oblige: they do their best to give each couple a valid pair of numbers such that their maximum is minimal.
But some rangers, among their numbers Byteasar, grew tired of this extracurricular activity and ensuing puzzles. Byteasar has asked you to write a program that will ease the process. The program, given a description of the cave and the current location of \(k\) couples, should determine \(k\) pairs of numbers \(x_i\) and \(y_i\) such that if the \(i\)-th couple follows respectively: he \(x_i\) and she \(y_i\) corridors marked with arrows,then they will meet in a single chamber of the cave \(max(x_i,y_i)\) is minimal,subject to above \(min(x_i,y_i)\) is minimal,if above conditions do not determine a unique solution, then the woman should cover smaller distance (\(x_i\ge y_i\)).
It may happen that such numbers \(x_i\) and \(y_i\) do not exist - then let \(x_i=y_i=-1\). Note that it is fine for several couples to meet in a single chamber. Once the lovers have found their dates, they will be happy to lose themselves in the cave again...
给定一棵内向森林，多次给定两个点a和b，求点对(x,y)满足：
1.从a出发走x步和从b出发走y步会到达同一个点
2.在1的基础上如果有多解，那么要求max(x,y)最小
3.在1和2的基础上如果有多解，那么要求min(x,y)最小
4.如果在1、2、3的基础上仍有多解，那么要求x>=y
输入格式
In the first line of the standard input there are two positive integers \(n\) and \(k\)(\(1\le n,k\le 500\ 000\)), separated by a single space, that denote the number of chambers in the Arrow Cave and the number of couples who want to find their dates, respectively.
The chambers are numbered from 1 to \(n\), while the enamoured couples are numbered from 1 to \(k\).
The second line of input contains \(n\) positive integers separated by single spaces:
the \(i\)-th such integer determines the number of chamber to which the corridor marked with an arrow going out of chamber \(i\) leads.
The following \(k\) lines specify the queries by the separated couples. Each such query consists of two positive integers separated by a single space - these denote the numbers of chambers where the lovers are - first him, then her.
In the tests worth 40% of the total points it additionally holds that \(n,k\le 2\ 000\).
输出格式
Your program should print exactly \(k\) lines to the standard output, one line per each couple specified in the input:
the \(i\)-th line of the output should give the instructions for the \(i\)-th couple on the input.
I.e., the \(i\)-th line of output should contain the integers \(x_i,y_i\), separated by a single space.
样例 #1
样例输入 #1
12 5
4 3 5 5 1 1 12 12 9 9 7 1
7 2
8 11
1 2
9 10
10 5
样例输出 #1
2 3
1 2
2 2
0 1
-1 -1
提示
给定一棵内向基环森林，多次给定两个点a和b，求点对(x,y)满足：
1.从a出发走x步和从b出发走y步会到达同一个点
2.在1的基础上如果有多解，那么要求max(x,y)最小
3.在1和2的基础上如果有多解，那么要求min(x,y)最小
4.如果在1、2、3的基础上仍有多解，那么要求x>=y
看看样例，其实就能发现，这是一个内向基环树森林，基环树森林。。自环也算环哦。。。
对于基环树，其实直接缩点，森林也是很好处理。但是问题在于是5e5个询问，同时还需要处理多解的情况。
先考虑怎么求出一个解吧
其实只需要管\(x-y\)的值即可（x-y指步数之差的绝对值），先求出lca，如果他们的lca是经过缩短的基环，由于是单向边的原因，依旧是一个确定的解，除非是围绕着这个环再转一圈。多解的情况感觉非常少，因为是单向树的原因，找到lca后，这两个点只能一起走动，而如果他们之间的距离差减去lca产生的距离差不为零，那么只能通过绕环走一圈来消除。也就是这个距离差减去lca产生的距离差，这个值如果不是环的长度的整数倍，那么就是确定无解。
哦，看错题了。拿这题没了啊，直接缩点，求lca，要么无解，要么lca不在基环上，要么lca在基环上。其中第一个情况，是很好判断的。直接看看是否在同一棵树上，第二个，这个点对就是两个点到lca的距离，看看要求，\(max(x,y)\)最小，如果向上继续的话，两个都变大。
而第三个情况，其实也是，直接让走到基环上后，小的步数走到另一个点的位置即可。后面的条件是不是有点多余？
不知道啊。先写吧
写到屎了
250行，70tps，实在找不到错了，写了一天了。
不管了，反正该学的学到了
就这样吧
#include<bits/stdc++.h>
#define ll long long
using namespace std;
inline int read(){
int a=0,b=1;char c=getchar();
for(;c<'0'||c>'9';c=getchar())if(c=='-')b=-1;
for(;c>='0'&&c<='9';c=getchar())a=a*10+c-'0';
return a*b;
}
struct edge
{
int next,to;
}e1[500001],e2[1000001];
int head1[500001],head2[1000001],tot1,tot2,n,k;
inline void add1(int i,int j)
{
e1[++tot1].next=head1[i];
e1[tot1].to=j;
head1[i]=tot1;
}
inline void add2(int i,int j)
{
e2[++tot2].next=head2[i];
e2[tot2].to=j;
head2[i]=tot2;
}
bool vis[500001];int deg[500001],a[500001];
queue<int> q;
void topo()
{
for(int i=1;i<=n;i++)
{
if(deg[i]==0)q.push(i);
}
while(!q.empty())
{
int x=q.front();
q.pop();
for(int i=head1[x];i!=0;i=e1[i].next)
{
int u=e1[i].to;
// if(deg[u]==0)continue;
deg[u]--;
if(deg[u]==0)
{
q.push(u);
}
}
}
}
vector<int> rmap[500001];
map<int,int> ma;
int bcj[500001][2];
int Get_fa(int x,int k)
{
if(bcj[x][k]==x)return x;
return bcj[x][k]=Get_fa(bcj[x][k],k);
}
int cnt=0;
vector<int> root;
void tarjan_awa(int st)
{
q.push(st);
root.push_back(++cnt);
while(!q.empty())
{
int x=q.front();
q.pop();
if(!deg[x])continue;
if(vis[x])continue;
vis[x]=1;
rmap[cnt].push_back(x);
ma[x]=cnt;
for(int i=head1[x];i!=0;i=e1[i].next)
{
int u=e1[i].to;
if(!deg[u])continue;
q.push(u);
}
}
}
int dep[500001][21],f[500001][21];
int num[500001];
void dfs(int x,int fa,int now)
{
vis[x]=1;
// cout << x <<endl;
dep[x][0]=now-1;f[x][0]=fa;
for(int i=head2[x];i!=0;i=e2[i].next)
{
int u=e2[i].to;
if(vis[u])continue;
if(u==fa)continue;
dfs(u,x,now+1);
}
}
int ask_lca(int x,int y)
{
if(dep[x][0]<dep[y][0])swap(x,y);
int now=19;
while(now>=0)
{
if(dep[x][now]>dep[y][0])x=f[x][now];
now--;
}
now=19;
if(x==y)return x;
while(now>=0)
{
if(f[x][now]!=f[y][now])y=f[y][now],x=f[x][now];
now--;
}
return f[x][0];
//mty dzs¿ìȥдÌâ ËäÈ»ÎÒ²ÂÄã¿´²»µ½
}
void Get_sequence(int x,int fa,int now)
{
num[x]=++now;
for(int i=head1[x];i!=0;i=e1[i].next)
{
int u=e1[i].to;
if(deg[u]==0||num[u]!=0)continue;
Get_sequence(u,x,now);
}
}
int main()
{
n=read();k=read();
for(int i=1;i<=n;i++)bcj[i][0]=bcj[i][1]=i;
for(int i=1;i<=n;i++)
{
int x=read();a[i]=x;
add1(i,x);deg[x]++;
bcj[i][0]=Get_fa(x,0);
}
topo();
for(int i=1;i<=n;i++)
{
if(deg[i]==0)bcj[i][1]=Get_fa(a[i],1);
}
// for(int i=1;i<=n;i++)
// {
// cout<<Get_fa(i,1)<<' ';
// }
// cout<<endl;
for(int i=1;i<=n;i++)
{
if(deg[i]!=0&&num[i]==0)
{
Get_sequence(i,0,0);
}
}
for(int i=1;i<=n;i++)//Ëõµã
{
if(deg[i]==0&&vis[i]==0)
{
vis[i]=1;
rmap[++cnt].push_back(i);
ma[i]=cnt;
}
else
if(deg[i]!=0&&vis[i]==0)
{
tarjan_awa(i);
}
}
// for(int i=1;i<=n;i++)
// {
// cout<<ma[i]<<' ';
// }
// cout<<endl;
for(int i=1;i<=n;i++)
{
add2(ma[i],ma[a[i]]);
add2(ma[a[i]],ma[i]);
}
memset(vis,0,sizeof(vis));
for(int i=0;i<root.size();i++)
{
// cout<<"?"<<endl;
dfs(root[i],0,1);
// cout<<root[i]<<' ';
}
// cout<<endl;
memset(vis,0,sizeof(vis));
for(int i=0;i<=18;i++)
{
for(int j=1;j<=cnt;j++)
{
dep[j][i+1]=dep[f[j][i]][i];
f[j][i+1]=f[f[j][i]][i];
}
}
// for(int i=1;i<=cnt;i++)
// {
// cout<<dep[i][0]<<' '<<f[i][0]<<endl;
// }
for(int i=1;i<=k;i++)
{
int x=read(),y=read();int rx=x,ry=y;
if(Get_fa(x,0)!=Get_fa(y,0))
{
cout<<-1<<' '<<-1<<endl;
continue;
}
x=ma[x],y=ma[y];
int lca=ask_lca(x,y);
if(rmap[lca].size()==1)
{
cout<<dep[x][0]-dep[lca][0]<<' '<<dep[y][0]-dep[lca][0]<<endl;
continue;
}
// cout<<lca<<endl;
int st1=Get_fa(rx,1),st2=Get_fa(ry,1);
st1=num[st1],st2=num[st2];
int movex,movey;
// cout<<st1<<' '<<st2<<endl;
if(st1==st2)movex=movey=0;
if(st1>st2)
{
movex=rmap[ma[Get_fa(rx,1)]].size()-st1+st2;
movey=st1-st2;
}
else
if(st1<st2)
{
movex=st2-st1;
movey=rmap[ma[Get_fa(rx,1)]].size()-st2+st1;
}
// cout<<dep[x][0]<<' '<<dep[y][0]<<' '<<movex<<' '<<movey<<endl;
int len1=dep[x][0],len2=dep[y][0];
// cout<<len1<<' '<<len2<<endl;
// if(max(len1+movex,len2)!=max(len2+movey,len1))
// {
if(max(len1+movex,len2)<=max(len2+movey,len1))
{
cout<<len1+movex<<' '<<len2<<endl;
}
else
{
cout<<len1<<' '<<len2+movey<<endl;
}
// }
// else
// {
// cout<<len1+movex<<' '<<len2<<endl;
// }
}
return 0;
}
/*
1 4
2 3
3 5
4 5
5 1
6 1
7 12
8 12
9 9
10 9
11 7
12 1
*/