P3398 仓鼠找 sugar（倍增法求LCA） · znzryb
## 思路讲解
判断树上a到b路径和c到d路径是否相交。
有一个小结论，就是如果两条路径相交，那么相交点至少是其中一条路径的最近公共祖先。
一呢，我们去构造反例，发现不太行，如果把那两个凸出来的点连接在一起就成环了。
那么我们发现如果不能连的话，相交点就成根节点了。
```cpp
// 保证共线
if(LCA(lcaAB,lcaCD)==lcaCD && LCA(c,lcaAB)==lcaAB){
cout<<"Y\n";
continue;
}
if(LCA(lcaAB,lcaCD)==lcaCD && LCA(lcaAB,d)==lcaAB){
cout<<"Y\n";
continue;
}
// ----------
// 看lcaCD在不在a->lcaAB上
if(LCA(lcaCD,lcaAB)==lcaAB && LCA(lcaCD,a)==lcaCD){
cout<<"Y\n";
continue;
}
if(LCA(lcaCD,lcaAB)==lcaAB && LCA(lcaCD,b)==lcaCD){
cout<<"Y\n";
continue;
}
```
## AC代码
[https://www.luogu.com.cn/record/217989209](https://www.luogu.com.cn/record/217989209)
::: 源代码
```cpp
// Problem: P3398 仓鼠找 sugar
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3398
// Memory Limit: 125 MB
// Time Limit: 1000 ms
// by znzryb
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define getFro(vec) (vec.empty()?0:vec.front())
#define getBac(vec) (vec.empty()?0:vec.front())
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define DEBUG(variable) \
do { \
std::cerr << #variable << ":"; \
for (const auto& elem : variable) { \
std::cerr << elem << " "; \
} \
std::cerr << "\n"; \
} while (0)
#define uniVec(var) \
do { \
sort(var.begin(),var.end());\
var.resize(unique(var.begin(),var.end())-var.begin());\
} while (0)
#define debugN(var,N) \
do{ \
std::cerr<<#var<<":"; \
FOR(i,1,N){ \
std::cerr<<var[i]<<" "; \
} \
std::cerr<<"\n"; \
}while(0)
#define debugMap(variable) \
do { \
std::cerr << #variable << ":\n"; \
for (const auto& pair : variable) { \
std::cerr << " " << pair.first << " => " << pair.second << "\n"; \
} \
std::cerr << std::endl; \
} while (0)
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef __int128 i128;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef long double LD;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(1e5)+10,INF=static_cast<ll>(1e18)+3;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;
ll N,M,Q,K,T;
ll Deep[MAXN],Anc[MAXN][30];
bool vis[MAXN];
/*
*/
vector<ll> g[MAXN];
void dfs(ll x,ll fa){
if(vis[x]) return;
vis[x]=true;
Deep[x]=Deep[fa]+1;
Anc[x][0]=fa;
FOR(i,0,SZ(g[x])-1){
ll vnode=g[x][i];
if(vis[vnode]) continue;
dfs(vnode,x);
}
}
inline ll LCA(ll a,ll b){ // 倍增法求LCA
if(Deep[a]<Deep[b]) swap(a,b);
ROF(i,18,0){
if(Deep[Anc[a][i]]>=Deep[b]){
a=Anc[a][i];
}
}
if(a==b) return a;
ROF(i,18,0){
if(Anc[a][i]!=Anc[b][i]){
a=Anc[a][i],b=Anc[b][i];
}
}
return Anc[a][0];
}
inline void solve(){
cin>>N>>Q;
FOR(i,1,N-1){
ll u,v;
cin>>u>>v;
g[u].pb(v);
g[v].pb(u);
}
dfs(1,0);
FOR(j,1,18){
FOR(i,1,N){
Anc[i][j]=Anc[Anc[i][j-1]][j-1];
}
}
FOR(i,1,Q){
ll a,b,c,d;
cin>>a>>b>>c>>d;
if(Deep[a]<Deep[b]) swap(a,b);
if(Deep[c]<Deep[d]) swap(c,d);
ll lcaAB=LCA(a,b);
ll lcaCD=LCA(c,d);
// if(Deep[lcaAB] Deep[lcaCD]){
// swap(a,c);
// swap(b,d);
// }
// 保证共线
if(LCA(lcaAB,lcaCD)==lcaCD && LCA(c,lcaAB)==lcaAB){
cout<<"Y\n";
continue;
}
if(LCA(lcaAB,lcaCD)==lcaCD && LCA(lcaAB,d)==lcaAB){
cout<<"Y\n";
continue;
}
// ----------
// 看lcaCD在不在a->lcaAB上
if(LCA(lcaCD,lcaAB)==lcaAB && LCA(lcaCD,a)==lcaCD){
cout<<"Y\n";
continue;
}
if(LCA(lcaCD,lcaAB)==lcaAB && LCA(lcaCD,b)==lcaCD){
cout<<"Y\n";
continue;
}
cout<<"N\n";
// || LCA(lcaAB,c)==lcaAB || LCA(lcaAB,d)==lcaAB
// if(Deep[lca]>=Deep[b] && Deep[lca]>=Deep[d]){
// cout<<"Y\n";
// }else{
// cout<<"N\n";
// }
}
// #ifdef LOCAL
// FOR(i,1,N){
// FOR(j,0,18){
// cerr<<Anc[i][j]<<" ";
// }
// cerr<<"\n";
// }
// #endif
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
// cin>>T;
// while(T--){
// solve();
// }
solve();
return 0;
}
/*
AC https://www.luogu.com.cn/record/217989209
*/
```
:::
## 心路历程（WA，TLE，MLE……）
WA
[https://www.luogu.com.cn/record/217981621](https://www.luogu.com.cn/record/217981621)
主要是无法处理这种情况。
::: 源代码
```cpp
// Problem: P3398 仓鼠找 sugar
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3398
// Memory Limit: 125 MB
// Time Limit: 1000 ms
// by znzryb
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define getFro(vec) (vec.empty()?0:vec.front())
#define getBac(vec) (vec.empty()?0:vec.front())
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define DEBUG(variable) \
do { \
std::cerr << #variable << ":"; \
for (const auto& elem : variable) { \
std::cerr << elem << " "; \
} \
std::cerr << "\n"; \
} while (0)
#define uniVec(var) \
do { \
sort(var.begin(),var.end());\
var.resize(unique(var.begin(),var.end())-var.begin());\
} while (0)
#define debugN(var,N) \
do{ \
std::cerr<<#var<<":"; \
FOR(i,1,N){ \
std::cerr<<var[i]<<" "; \
} \
std::cerr<<"\n"; \
}while(0)
#define debugMap(variable) \
do { \
std::cerr << #variable << ":\n"; \
for (const auto& pair : variable) { \
std::cerr << " " << pair.first << " => " << pair.second << "\n"; \
} \
std::cerr << std::endl; \
} while (0)
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef __int128 i128;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef long double LD;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(1e5)+10,INF=static_cast<ll>(1e18)+3;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;
ll N,M,Q,K,T;
ll Deep[MAXN],Anc[MAXN][30];
bool vis[MAXN];
/*
*/
vector<ll> g[MAXN];
void dfs(ll x,ll fa){
if(vis[x]) return;
vis[x]=true;
Deep[x]=Deep[fa]+1;
Anc[x][0]=fa;
FOR(i,0,SZ(g[x])-1){
ll vnode=g[x][i];
if(vis[vnode]) continue;
dfs(vnode,x);
}
}
inline ll LCA(ll a,ll b){ // 倍增法求LCA
if(Deep[a]<Deep[b]) swap(a,b);
ROF(i,18,0){
if(Deep[Anc[a][i]]>=Deep[b]){
a=Anc[a][i];
}
}
if(a==b) return a;
ROF(i,18,0){
if(Anc[a][i]!=Anc[b][i]){
a=Anc[a][i],b=Anc[b][i];
}
}
return Anc[a][0];
}
inline void solve(){
cin>>N>>Q;
FOR(i,1,N-1){
ll u,v;
cin>>u>>v;
g[u].pb(v);
g[v].pb(u);
}
dfs(1,0);
FOR(j,1,18){
FOR(i,1,N){
Anc[i][j]=Anc[Anc[i][j-1]][j-1];
}
}
FOR(i,1,Q){
ll a,b,c,d;
cin>>a>>b>>c>>d;
if(Deep[a]<Deep[b]) swap(a,b);
if(Deep[c]<Deep[d]) swap(c,d);
ll lcaAB=LCA(a,b);
ll lcaCD=LCA(c,d);
// if(Deep[lcaAB] Deep[lcaCD]){
// swap(a,c);
// swap(b,d);
// }
// 保证共线
if(LCA(a,lcaCD)==lcaCD && LCA(c,a)==a){
cout<<"Y\n";
continue;
}
if(LCA(b,lcaCD)==lcaCD && LCA(c,b)==b){
cout<<"Y\n";
continue;
}
if(LCA(a,lcaCD)==lcaCD && LCA(d,a)==a){
cout<<"Y\n";
continue;
}
if(LCA(b,lcaCD)==lcaCD && LCA(d,b)==b){
cout<<"Y\n";
continue;
}
// ----------
if(LCA(c,lcaAB)==lcaAB && LCA(c,a)==c){
cout<<"Y\n";
continue;
}
if(LCA(d,lcaAB)==lcaAB && LCA(d,a)==d){
cout<<"Y\n";
continue;
}
if(LCA(c,lcaAB)==lcaAB && LCA(c,b)==c){
cout<<"Y\n";
continue;
}
if(LCA(d,lcaAB)==lcaAB && LCA(d,b)==d){
cout<<"Y\n";
continue;
}
cout<<"N\n";
// || LCA(lcaAB,c)==lcaAB || LCA(lcaAB,d)==lcaAB
// if(Deep[lca]>=Deep[b] && Deep[lca]>=Deep[d]){
// cout<<"Y\n";
// }else{
// cout<<"N\n";
// }
}
// #ifdef LOCAL
// FOR(i,1,N){
// FOR(j,0,18){
// cerr<<Anc[i][j]<<" ";
// }
// cerr<<"\n";
// }
// #endif
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
// cin>>T;
// while(T--){
// solve();
// }
solve();
return 0;
}
/*
*/
```
:::