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
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]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>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>1;
int ret1 = querymin(o<<1,tl,mid,l,r);
int ret2 = querymin(o<<1|1,mid+1,tr,l,r);
if(in[ret1]>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 G[maxn];
vector 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 q;
q.push(1);
dep[1] = 1;
while(!q.empty()){
int u = q.front();
q.pop();
for(int i=0;idep[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;iin[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]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
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]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>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>1;
int ret1 = querymin(o<<1,tl,mid,l,r);
int ret2 = querymin(o<<1|1,mid+1,tr,l,r);
if(in[ret1]>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 G[maxn];
vector 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;iy)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;iin[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]dd[x2]){
resm = v0;
res = dd[x1];
}
else{
resm = u0;
res = dd[x2];
}
printf("%d %d\n",resm,res-1);
}
return 0;
}