链接：http://codeforces.com/contest/1062/problem/E
思路：求一个区间内删去一个点后深度最深的lca，首先我们要了解与一个性质，如果按照dfs序下来，那么一个区间内的lca就等于dfs区间中dfs序最小和最大的两个点的lca（其他的点一定都属于这两个点lca的子树里面，可自行证明），这个性质是我们做这个题的前提题，然后区间最值的维护方法有很多，这里我用了线段树不过显得麻烦了，简单一点可以用st表，首先选出最大点u和最小点v，去除最大掉就在（l,u-1）U (u+1,r)中寻找最大值与v求lca，同理对于去除v，最终比较一下两种情况下谁的lca的深度最深就是答案了。顺便用这个题练习了一下倍增法求lca以及lca转RMQ问题，tarjan离线在这个题上我还不太清楚怎么使用，反正通过这个题把三个求lca的板子都顺带写了一下，至于树剖的版本等以后学了再补上来吧。
倍增法版本：
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+100;
int n,q,t;
int maxv[maxn<<2],minv[maxn<<2],mini[maxn<<2],maxi[maxn<<2];
int in[maxn];
void pushup(int o){
if(minv[o<<1]<minv[o<<1|1]){
minv[o] = minv[o<<1];
mini[o] = mini[o<<1];
}
else{
minv[o] = minv[o<<1|1];
mini[o] = mini[o<<1|1];
}
if(maxv[o<<1]>maxv[o<<1|1]){
maxv[o] = maxv[o<<1];
maxi[o] = maxi[o<<1];
}
else{
maxv[o] = maxv[o<<1|1];
maxi[o] = maxi[o<<1|1];
}
}
void build(int o,int l,int r){
if(l<r){
int mid = l+r>>1;
build(o<<1,l,mid);
build(o<<1|1,mid+1,r);
pushup(o);
}
else{
minv[o] = maxv[o] = in[l];
mini[o] = l;
maxi[o] = l;
}
}
int querymin(int o,int tl,int tr,int l,int r){
if(tr<l||r<tl)return maxn-1;
if(l<=tl&&tr<=r)return mini[o];
int mid = (tl+tr)>>1;
int ret1 = querymin(o<<1,tl,mid,l,r);
int ret2 = querymin(o<<1|1,mid+1,tr,l,r);
if(in[ret1]<in[ret2])return ret1;
return ret2;
}
int querymax(int o,int tl,int tr,int l,int r){
if(tr<l||r<tl)return maxn-2;
if(l<=tl&&tr<=r)return maxi[o];
int mid = (tl+tr)>>1;
int ret1 = querymax(o<<1,tl,mid,l,r);
int ret2 = querymax(o<<1|1,mid+1,tr,l,r);
if(in[ret1]>in[ret2])return ret1;
return ret2;
}
struct edge{
int from,to,dist;
};
struct LCA{
int f[maxn][25];
int d[maxn];
int dep[maxn];
vector<int> G[maxn];
vector<edge> edges;
int n,m,t;
void init(int n,int t){
this->n = n;
this->t = t;
for(int i=0;i<=n;i++)G[i].clear(),dep[i] = d[i] = 0;
edges.clear();
}
void addedge(int from,int to,int dist = 1){
edges.push_back(edge{from,to,dist});
m = edges.size();
G[from].push_back(m-1);
}
void bfs(){
queue<int> q;
q.push(1);
dep[1] = 1;
while(!q.empty()){
int u = q.front();
q.pop();
for(int i=0;i<G[u].size();i++){
edge e = edges[G[u][i]];
int v = e.to;
if(dep[v])continue;
dep[v] = dep[u] + 1;
d[v] = d[u] + e.dist;
f[v][0] = u;
for(int j=1;j<=t;j++){
f[v][j] = f[f[v][j-1]][j-1];
q.push(v);
}
}
}
}
int lca(int x,int y){
if(dep[x]>dep[y])swap(x,y);
for(int i=t;i>=0;i--){
if(dep[f[y][i]]>=dep[x])y = f[y][i];
}
if(x==y)return x;
for(int i=t;i>=0;i--){
if(f[x][i]!=f[y][i])x = f[x][i],y = f[y][i];
}
return f[x][0];
}
}solver;
void dfs(int u,int f){
in[u] = t++;
for(int i=0;i<solver.G[u].size();i++){
edge e = solver.edges[solver.G[u][i]];
int v = e.to;
if(v==f)continue;
dfs(v,u);
}
}
int main(){
scanf("%d%d",&n,&q);
solver.init(n,log(n)/log(2)+1);
for(int i=2;i<=n;i++){
int a;
scanf("%d",&a);
solver.addedge(i,a);
solver.addedge(a,i);
}
in[maxn-1] = 1e9;
in[maxn-2] = 0;
solver.bfs();
t = 1;
dfs(1,-1);
build(1,1,n);
// for(int i=1;i<=n;i++)printf("%d %d\n",i,in[i]);
for(int i=0;i<q;i++){
int l,r;
scanf("%d%d",&l,&r);
int u0 = querymax(1,1,n,l,r);
int v0 = querymin(1,1,n,l,r);
//printf("%d %d 111111111\n",u0,v0);
int uu,vv;
int u1 = querymax(1,1,n,l,u0-1);
int u2 = querymax(1,1,n,u0+1,r);
if(in[u1]>in[u2])uu = u1;
else uu = u2;
int v1 = querymin(1,1,n,l,v0-1);
int v2 = querymin(1,1,n,v0+1,r);
if(in[v1]<in[v2])vv = v1;
else vv = v2;
// printf("%d %d %d %d 3333333333\n",u1,u2,v1,v2);
//printf("%d %d\n",uu,vv);
int x1 = solver.lca(u0,vv);
int x2 = solver.lca(uu,v0);
int res = 0;
int resx = 1;
printf("%d %d\n",x1,x2);
//printf("%d %d %d %d 222222222222\n",x[0],x[1],x[2],x[3]);
//printf("%d %d %d %d 222222222222\n",solver.dep[x[0]],solver.dep[x[1]],solver.dep[x[2]],solver.dep[x[3]]);
int resm;
if(solver.dep[x1]>solver.dep[x2]){
resm = v0;
res = solver.dep[x1];
}
else{
resm = u0;
res = solver.dep[x2];
}
printf("%d %d\n",resm,res-1);
}
return 0;
}
RMQ版本
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+100;
int n,q,t;
int maxv[maxn<<2],minv[maxn<<2],mini[maxn<<2],maxi[maxn<<2];
int in[maxn];
void pushup(int o){
if(minv[o<<1]<minv[o<<1|1]){
minv[o] = minv[o<<1];
mini[o] = mini[o<<1];
}
else{
minv[o] = minv[o<<1|1];
mini[o] = mini[o<<1|1];
}
if(maxv[o<<1]>maxv[o<<1|1]){
maxv[o] = maxv[o<<1];
maxi[o] = maxi[o<<1];
}
else{
maxv[o] = maxv[o<<1|1];
maxi[o] = maxi[o<<1|1];
}
}
void build(int o,int l,int r){
if(l<r){
int mid = l+r>>1;
build(o<<1,l,mid);
build(o<<1|1,mid+1,r);
pushup(o);
}
else{
minv[o] = maxv[o] = in[l];
mini[o] = l;
maxi[o] = l;
}
}
int querymin(int o,int tl,int tr,int l,int r){
if(tr<l||r<tl)return maxn-1;
if(l<=tl&&tr<=r)return mini[o];
int mid = (tl+tr)>>1;
int ret1 = querymin(o<<1,tl,mid,l,r);
int ret2 = querymin(o<<1|1,mid+1,tr,l,r);
if(in[ret1]<in[ret2])return ret1;
return ret2;
}
int querymax(int o,int tl,int tr,int l,int r){
if(tr<l||r<tl)return maxn-2;
if(l<=tl&&tr<=r)return maxi[o];
int mid = (tl+tr)>>1;
int ret1 = querymax(o<<1,tl,mid,l,r);
int ret2 = querymax(o<<1|1,mid+1,tr,l,r);
if(in[ret1]>in[ret2])return ret1;
return ret2;
}
struct edge{
int from,to,dist;
};
struct LCA{
int n,m;
vector<int> G[maxn];
vector<edge> edges;
int ntime;
int dep[maxn<<1];
int d[maxn<<1][25];
int first[maxn];
int vis[maxn<<1];
void init(int n){
this->n = n;
for(int i=0;i<=n;i++)G[i].clear();
edges.clear();
ntime = 1;
memset(vis,0,sizeof(vis));
memset(first,0,sizeof(first));
memset(dep,0,sizeof(dep));
}
void addedge(int from,int to,int dist = 1){
edges.push_back(edge{from,to,dist});
m = edges.size();
G[from].push_back(m-1);
}
void dfs(int u,int f,int d){
first[u] = ntime;
vis[ntime] = u;
dep[ntime++] = d;
for(int i=0;i<G[u].size();i++){
edge e = edges[G[u][i]];
int v = e.to;
if(v==f)continue;
dfs(v,u,d+1);
vis[ntime] = u;
dep[ntime++] = d;
}
}
void RMQ_init(int len){
for(int i=1;i<=len;i++)d[i][0] = i;
for(int j=1;(1<<j)<=len;j++){
for(int i=1;i+(1<<j)-1<=len;i++){
int a = d[i][j-1];
int b = d[i+(1<<(j-1))][j-1];
if(dep[a]<=dep[b])d[i][j] = a;
else d[i][j] = b;
}
}
}
int query(int l,int r){
int k = 0;
while((1<<(k+1))<=r-l+1)k++;
int a = d[l][k];
int b = d[r-(1<<k)+1][k];
if(dep[a]<=dep[b])return a;
return b;
}
int lca(int u,int v){
int x = first[u];
int y = first[v];
if(x>y)swap(x,y);
return vis[query(x,y)];
}
void solve(){
dfs(1,-1,0);
RMQ_init(ntime-1);
}
}solver;
int dd[maxn];
void dfs(int u,int f){
in[u] = t++;
for(int i=0;i<solver.G[u].size();i++){
edge e = solver.edges[solver.G[u][i]];
int v = e.to;
if(v==f)continue;
dd[v] = dd[u] + 1;
dfs(v,u);
}
}
int main(){
scanf("%d%d",&n,&q);
solver.init(n);
for(int i=2;i<=n;i++){
int a;
scanf("%d",&a);
solver.addedge(i,a);
solver.addedge(a,i);
}
solver.solve();
in[maxn-1] = 1e9;
in[maxn-2] = 0;
t = 1;
dd[1] = 1;
dfs(1,-1);
build(1,1,n);
//for(int i=1;i<=n;i++)printf("%d %d\n",i,dd[i]);
for(int i=0;i<q;i++){
int l,r;
scanf("%d%d",&l,&r);
int u0 = querymax(1,1,n,l,r);
int v0 = querymin(1,1,n,l,r);
//printf("%d %d 111111111\n",u0,v0);
int uu,vv;
int u1 = querymax(1,1,n,l,u0-1);
int u2 = querymax(1,1,n,u0+1,r);
if(in[u1]>in[u2])uu = u1;
else uu = u2;
int v1 = querymin(1,1,n,l,v0-1);
int v2 = querymin(1,1,n,v0+1,r);
if(in[v1]<in[v2])vv = v1;
else vv = v2;
//printf("%d %d %d %d 3333333333\n",u1,u2,v1,v2);
//printf("%d %d\n",uu,vv);
int x1 = solver.lca(u0,vv);
int x2 = solver.lca(uu,v0);
int res = 0;
int resx = 1;
//printf("%d %d\n",x1,x2);
//printf("%d %d %d %d 222222222222\n",x[0],x[1],x[2],x[3]);
//printf("%d %d %d %d 222222222222\n",solver.dep[x[0]],solver.dep[x[1]],solver.dep[x[2]],solver.dep[x[3]]);
int resm;
if(dd[x1]>dd[x2]){
resm = v0;
res = dd[x1];
}
else{
resm = u0;
res = dd[x2];
}
printf("%d %d\n",resm,res-1);
}
return 0;
}
