标签：更新 mat connect include wing nta memset case 有一个
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.
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).
For each case,output Q integers means the LCA of [li,ri].
5
1 2
1 3
3 4
4 5
5
1 2
2 3
3 4
3 5
1 51
1
3
3
1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;
}标签：更新 mat connect include wing nta memset case 有一个
原文地址：https://www.cnblogs.com/ezoihy/p/9112927.html
