CF983E NN country

题目

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题目描述
In the NN country, there are nn cities, numbered from 11 to nn , and n - 1n−1 roads, connecting them. There is a roads path between any two cities.

There are mm bidirectional bus routes between cities. Buses drive between two cities taking the shortest path with stops in every city they drive through. Travelling by bus, you can travel from any stop on the route to any other. You can travel between cities only by bus.

You are interested in qq questions: is it possible to get from one city to another and what is the minimum number of buses you need to use for it?

输入格式
The first line contains a single integer nn ( 2 \le n \le 2 \cdot 10^52≤n≤2⋅10
5
) — the number of cities.

The second line contains n - 1n−1 integers p_2, p_3, \ldots, p_np
2

,p
3

,…,p
n

( 1 \le p_i < i1≤p
i

<i ), where p_ip
i

means that cities p_ip
i

and ii are connected by road.

The third line contains a single integer mm ( 1 \le m \le 2 \cdot 10^51≤m≤2⋅10
5
) — the number of bus routes.

Each of the next mm lines contains 22 integers aa and bb ( 1 \le a, b \le n1≤a,b≤n , a \neq ba


=b ), meaning that there is a bus route between cities aa and bb . It is possible that there is more than one route between two cities.

The next line contains a single integer qq ( 1 \le q \le 2 \cdot 10^51≤q≤2⋅10
5
) — the number of questions you are interested in.

Each of the next qq lines contains 22 integers vv and uu ( 1 \le v, u \le n1≤v,u≤n , v \neq uv


=u ), meaning that you are interested if it is possible to get from city vv to city uu and what is the minimum number of buses you need to use for it.

输出格式
Print the answer for each question on a separate line. If there is no way to get from one city to another, print -1−1 . Otherwise print the minimum number of buses you have to use.

输入输出样例
输入 #1复制
7
1 1 1 4 5 6
4
4 2
5 4
1 3
6 7
6
4 5
3 5
7 2
4 5
3 2
5 3
输出 #1复制
1
3
-1
1
2
3
输入 #2复制
7
1 1 2 3 4 1
4
4 7
3 5
7 6
7 6
6
4 6
3 1
3 2
2 7
6 3
5 3
输出 #2复制
1
-1
-1
1
-1
1
说明/提示
Routes for first sample are marked on the picture.

思路

发现可以贪心，贪心可以倍增优化。

贪心策略：把路径拆开成两条链，起始点和结尾点为祖先后代关系的链，显然是深度越浅越好

然后就可以暴力爬。发现一种特殊情况：有路径可以穿过lca

那么就判一下，是否有路径起始点在（倍增跳过后的停留点）子树中。

如何判？子树中的点的dfs序一定 in[rt]<=x<=out[rt]

两个变量就是二维数点问题咯。。

代码

#include<bits/stdc++.h>
using namespace std;
const int N=2e5+77;
struct edge{int link,next;
}e[N<<1];
struct event{int x,y,id,opt;
}b[N*5];
int m,q,n,head[N],tot;
void add_edge(int u,int v){e[++tot]=(edge){v,head[u]}; head[u]=tot;
}
void insert(int u,int v){add_edge(u,v); add_edge(v,u);
}
int g[N][20],dep[N],fa[N][20],l[N],r[N],cnt;
void dfs(int u,int Fa){dep[u]=dep[Fa]+1; fa[u][0]=Fa; l[u]=++cnt;for (int i=1;i<20;i++) fa[u][i]=fa[fa[u][i-1]][i-1];for (int i=head[u];i;i=e[i].next){int v=e[i].link;if (v!=Fa) dfs(v,u);}r[u]=cnt;
}
int LCA(int u,int v){if (dep[u]<dep[v]) swap(u,v);int delta=dep[u]-dep[v];for (int i=0;i<20;i++) if (delta&(1<<i)) u=fa[u][i];for (int i=19;i>=0;i--) if (fa[u][i]!=fa[v][i]) u=fa[u][i],v=fa[v][i];if (u!=v) return fa[u][0];return u;
}
void init(){n=read();for (int i=2;i<=n;i++) insert(i,read());dfs(1,0); m=read(); tot=0;for (int i=1;i<=m;i++){int u=read(),v=read(),lca=LCA(u,v);if (l[u]>l[v]) swap(u,v);if (!g[u][0]||dep[lca]<dep[g[u][0]]) g[u][0]=lca;if (!g[v][0]||dep[lca]<dep[g[v][0]]) g[v][0]=lca;b[++tot]=(event){l[u],l[v],0,1};}
}
void Dfs(int u,int Fa){for (int i=head[u];i;i=e[i].next){int v=e[i].link;if (v!=Fa){Dfs(v,u);if (!g[u][0]||g[v][0]&&dep[g[v][0]]<dep[g[u][0]]) g[u][0]=g[v][0];}}
}
bool cmp(event A,event B){return A.x<B.x||(A.x==B.x&&A.id<B.id);
}
int bit[N];
int lowbit(int x){return x&(-x);
}
void update(int x){for (;x<=n;x+=lowbit(x)) bit[x]++;
}
int query(int x){int sum=0;for (;x;x-=lowbit(x)) sum+=bit[x];return sum;
}
int ans[N],ANS[N];
void solve(){Dfs(1,0);for (int i=1;i<=n;i++) if (g[i][0]==i) g[i][0]=0;for (int i=1;i<20;i++)for (int j=1;j<=n;j++) g[j][i]=g[g[j][i-1]][i-1];q=read();for (int i=1;i<=q;i++){int u=read(),v=read(),lca=LCA(u,v);if (l[u]>l[v]) swap(u,v);for (int j=19;j>=0;j--) if (g[u][j]&&dep[g[u][j]]>dep[lca]) u=g[u][j],ans[i]+=(1<<j);for (int j=19;j>=0;j--) if (g[v][j]&&dep[g[v][j]]>dep[lca]) v=g[v][j],ans[i]+=(1<<j);if ((!g[u][0]&&u!=lca)||(!g[v][0]&&v!=lca)) {ans[i]=-1; continue;}if (u==lca||v==lca) ans[i]++;else {ans[i]+=2;b[++tot]=(event){r[u],r[v],i,1};b[++tot]=(event){l[u]-1,r[v],i,-1};b[++tot]=(event){r[u],l[v]-1,i,-1};b[++tot]=(event){l[u]-1,l[v]-1,i,1};}}sort(b+1,b+1+tot,cmp);for (int i=1;i<=tot;i++){if (!b[i].id) update(b[i].y);else ANS[b[i].id]+=b[i].opt*query(b[i].y);}for (int i=1;i<=q;i++) writeln(ans[i]-(ANS[i]>0));
}
int main(){init(); solve();return 0;
}

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CF983E NN country

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