HDU5266 LCA

Description

给一棵 n 个点的树，Q 个询问 [L,R] : 求点 L , 点 L+1 , 点 L+2 …… 点 R 的 LCA.

Input

多组数据.

The following line contains an integers,n(2≤n≤300000).

AT The following n−1 line, two integers are bi and ci at every line, it shows an edge connecting bi and ci.

The following line contains ans integers,Q(Q≤300000).

AT The following Q line contains two integers li and ri(1≤li≤ri≤n).

Output

For each case,output Q integers means the LCA of [li,ri].

Sample Input

5
1 2
1 3
3 4
4 5
5
1 2
2 3
3 4
3 5
1 5

Sample Output

1
1
3
3
1

solution

这题其实就是求[l,r]区间内的公共lca。

既,
\[ans(l,r)=LCA(a_l,a_{l+1},a_{l+2}\cdots a_r)\]
这里有一个显而易见的结论
\[LCA(x,y,z)=LCA\bigl(LCA(x,y),z\bigr)=LCA\bigl(LCA(z,y),x\bigr)=LCA\bigl(LCA(z,x),y\bigr)\]

所以我们在这里考虑建一棵线段树，每次pushup向上更新lca，我们可以用树剖来求lca，这样我们就可以求出区间lca了

这是代码

#include<iostream>
#include<cstdio>
#include<cstring>using namespace std;
const int maxn=300001,inf=0x7fffffff;
int n,m,tot,root,nxt[maxn<<1],to[maxn<<1],head[maxn],lca[maxn<<2],dep[maxn],siz[maxn],top[maxn],fa[maxn],son[maxn];
bool check[maxn];void addedge(int x,int y){nxt[++tot]=head[x];head[x]=tot;to[tot]=y;
}void dfs1(int u,int f) {dep[u]=dep[fa[u]=f]+(siz[u]=1);for(int i=head[u];i;i=nxt[i]) {int v=to[i];if(v==f)continue;dfs1(v,u);siz[u]+=siz[v];if(siz[v]>siz[son[u]])son[u]=v;}
}void dfs2(int u,int topf){top[u]=topf;if(!son[u])return;dfs2(son[u],topf);for(int i=head[u];i;i=nxt[i]){int v=to[i];if(v==fa[u] or v==son[u])continue;dfs2(v,v);}
}int Lca(int x,int y) {register int u=x,v=y;while(top[u]!=top[v]) {if(dep[top[u]]<dep[top[v]])swap(u,v);u=fa[top[u]];}return dep[u]<=dep[v]?u:v;
}void pushup(int o){lca[o]=Lca(lca[o<<1],lca[o<<1|1]);
}void build(int o,int l,int r){if(l==r){lca[o]=l;return;}int mid=l+r>>1;build(o<<1,l,mid);build(o<<1|1,mid+1,r);pushup(o);
}int query(int o,int l,int r,int x,int y){if(x<=l and r<=y){return lca[o];}int mid=l+r>>1,ans1=-1,ans2=-1;if(x<=mid)ans1=query(o<<1,l,mid,x,y);if(mid+1<=y)ans2=query(o<<1|1,mid+1,r,x,y);if(ans1!=-1 and ans2!=-1)return Lca(ans1,ans2);if(ans1!=-1)return ans1;if(ans2!=-1)return ans2;
}int main(){while(~scanf("%d",&n)){tot=0;memset(fa,0,sizeof(fa));memset(son,0,sizeof(son));memset(head,0,sizeof(head));for(int i=1;i<n;i++){int u,v;scanf("%d%d",&u,&v);addedge(u,v),addedge(v,u);}dfs1(1,0);dfs2(1,1);build(1,1,n);scanf("%d",&m);while(m--){int l,r;scanf("%d%d",&l,&r);printf("%d\n",query(1,1,n,l,r));}}return 0;
}