Znzryb's Blog

P3919 【模板】可持久化线段树 1（可持久化数组）

Posted on 2025-05-22 Edited on 2026-02-17

思路讲解

其实不难，理解之后写起来还是挺简单的。

先是小T了一下，然后上了读写优化就好了。

AC代码

https://www.luogu.com.cn/record/218184451

源代码

// Problem: P3919 【模板】可持久化线段树 1（可持久化数组）
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3919
// Memory Limit: 1 MB
// Time Limit: 1500 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> arr3;
typedef array<ll,2> arr2;
typedef double DB;
typedef long double LD;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(1e6)+10,INF=static_cast<ll>(1e18)+3;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;

ll N,M,Q,X,K,T,A[MAXN];
/*

*/
struct zhuXiTree{	// 主席树单点修改
	ll Cver=0;
	struct Tree{
		ll lson,rson,val,l,r;
	};
	ll idx=1;
	vector<Tree> tr;vector<ll> verN,org;
	zhuXiTree(ll n,ll m){	// n为数组长度，m为查询次数
		ll ln=n*4+m*ceil(log2(n))+131;
		tr.resize(ln);
		verN.resize(m+5);
		org.resize(n+5);
		verN[Cver]=idx;
	}
	void buildTree(ll node,ll l,ll r){
		if(l==r){
			tr[node].val=org[l];
			tr[node].l=l;
			tr[node].r=r;
			return;
		}
		tr[node].l=l;
		tr[node].r=r;
		tr[node].lson=++idx;
		tr[node].rson=++idx;
		buildTree(tr[node].lson,l,(l+r)>>1);
		buildTree(tr[node].rson,((l+r)>>1)+1,r);
	}
	void modify(ll node,ll ver,ll pos,ll x){
		tr[++idx].lson=tr[node].lson;
		tr[idx].rson=tr[node].rson;
		tr[idx].val=tr[node].val;
		tr[idx].l=tr[node].l;
		tr[idx].r=tr[node].r;
		if(verN[ver]==node){
			++Cver;
			verN[Cver]=idx;
		}
		if(tr[idx].l==tr[idx].r){
			if(tr[idx].l==pos) tr[idx].val=x;
			return;
		}
		ll mid=tr[idx].l+tr[idx].r>>1;
		if(pos<=mid){
			tr[idx].lson=idx+1;
			modify(tr[node].lson,ver,pos,x);
		}else{
			tr[idx].rson=idx+1;
			modify(tr[node].rson,ver,pos,x);
		}
	}
	ll query(ll node,ll ver,ll pos){
		tr[++idx].lson=tr[node].lson;
		tr[idx].rson=tr[node].rson;
		tr[idx].val=tr[node].val;
		tr[idx].l=tr[node].l;
		tr[idx].r=tr[node].r;
		if(verN[ver]==node){
			++Cver;
			verN[Cver]=idx;
		}
		if(tr[idx].l==tr[idx].r){
			return tr[idx].val;
		}
		ll mid=tr[idx].l+tr[idx].r>>1;
		if(pos<=mid){
			tr[idx].lson=idx+1;
			return query(tr[node].lson,ver,pos);
		}else{
			tr[idx].rson=idx+1;
			return query(tr[node].rson,ver,pos);
		}
	}
};

inline ll read()
{
	ll x=0,f=1;
	char ch=getchar();
	while(ch<'0'||ch>'9')
		{
			if(ch=='-')
				f=-1;
			ch=getchar();
		}
	while(ch>='0' && ch<='9')
		x=x*10+ch-'0',ch=getchar();
	return x*f;
}

inline void solve(){
	N=read();
	M=read();
	zhuXiTree tr(N,M);
	FOR(i,1,N){
		tr.org[i]=read();
	}
	tr.buildTree(1,1,N);
	FOR(i,1,M){
		ll ver=read(),op=read(),pos=read(),x;
		if(op==1){
			x=read();
			tr.modify(tr.verN[ver],ver,pos,x);
		}else{
			ll ans=tr.query(tr.verN[ver],ver,pos);
			cout<<ans<<"\n";
		}
	}
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	// cin>>T;
	// while(T--){
		// solve();
	// }
	solve();
	return 0;
}
/*
TLE https://www.luogu.com.cn/record/218183673 TLE了一个点
AC https://www.luogu.com.cn/record/218184451
*/

心路历程（WA，TLE，MLE……）

https://www.luogu.com.cn/record/218183673

TLE

ABC-405-F - Chord Crossing 和弦交叉（如何离线计算相交区间数量）

Posted on 2025-05-21 Edited on 2026-02-16

思路讲解

像这种题目，一般来说是通过一些离线想法，比如排序什么的，然后运用一些数据结构计算端点的数量。

这道题目本质上是要离线计算相交而不包含的区间数量。

通过离线，保证一个端点一定符合要求，另一个端点数量使用树状数组求出。

ll idx=1;
FOR(i,1,Q){
	ll l=que[i][0],r=que[i][1],ind=que[i][2];
	while(idx<=M){
		// 我们保证了实线r大于虚线r。
		if(lr[idx][1]<r){
			break;
		}
		trl.add(lr[idx][0],1);
		++idx;
	}
	// 要求实线l小于虚线r大于虚线l
	Ans[ind]=trl.query(l,r);
}
sort(que+1,que+1+Q);
sort(lr+1,lr+1+M);
idx=1;
FOR(i,1,Q){
	ll l=que[i][0],r=que[i][1],ind=que[i][2];
	while(idx<=M){
		// 我们保证了实线l小于虚线l。
		if(lr[idx][0]>l){
			break;
		}
		trr.add(lr[idx][1],1);
		++idx;
	}
	// 要求实线l小于虚线r大于虚线l
	Ans[ind]+=trr.query(l,r);
}

AC代码

https://atcoder.jp/contests/abc405/submissions/66033424

源代码

// Problem: F - Chord Crossing
// Contest: AtCoder - AtCoder Beginner Contest 405
// URL: https://atcoder.jp/contests/abc405/tasks/abc405_f
// Memory Limit: 1024 MB
// Time Limit: 3000 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> arr3;
typedef array<ll,2> arr2;
typedef double DB;
typedef long double LD;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(1e6)+10,INF=static_cast<ll>(1e18)+3;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;

ll N,M,K,T,A[MAXN];
ll Ans[MAXN];
arr2 lr[MAXN];
arr3 que[MAXN];
/*

*/
struct BIT {  // 普通的单点修改区间查询树状数组，含有查找第K个空位的功能
  vector<ll> tr;
  int n;
  inline int lowbit(int x) { return x & (-x); }
  BIT(int ln) {  // 构造+初始化函数
    n = ln;
    tr.resize(n+5,0);
  }
  inline void add(int p, int x) {
  	if(p==0) return;
    while (p <= n) {
      tr[p] += x;
      p += lowbit(p);
    }
  }
  inline ll query(int l, int r) {
    ll lres = 0, rres = 0;
    l -= 1;
    while (l > 0) {
      lres += tr[l];
      l -= lowbit(l);
    }
    while (r > 0) {
      rres += tr[r];
      r -= lowbit(r);
    }
    return rres - lres;
  }
  inline int findKth(int k) {  // 找到第k个空位
    int cx = 0;
    for (int i = 1 << 20; i > 0; i >>= 1) {  // 这个你可以理解为不断缩小步长的搜索
      if (cx + i <= n && lowbit(cx + i) - tr[cx + i] < k) {
        // lowbit(cx)-tr[cx+i] 表示cx+i这个树状数组区间有多少空位
        k -= lowbit(cx + i) - tr[cx + i];
        cx += i;
      }
    }
    return cx + 1;  // 我们始终让cx < k，所以返回的就是最大的比k小的，+1就是新空位出现的地方
  }
};
inline bool rFirst2(arr2 a,arr2 b){
	return a[1]>b[1];
}
inline bool rFirst3(arr3 a,arr3 b){
	return a[1]>b[1];
}
inline void solve(){
	cin>>N>>M;
	FOR(i,1,M){
		ll u,v;
		cin>>u>>v;
		lr[i]={u,v};
	}
	sort(lr+1,lr+1+M,rFirst2);
	ll Q;
	cin>>Q;
	FOR(i,1,Q){
		ll a,b;
		cin>>a>>b;
		que[i]={a,b,i};
	}
	// 按左端点从大到小，因为大的虚线左端点只用考虑比他还大的实线左端点
	sort(que+1,que+1+Q,rFirst3);
	BIT trl(2*N),trr(2*N);
	ll idx=1;
	FOR(i,1,Q){
		ll l=que[i][0],r=que[i][1],ind=que[i][2];
		while(idx<=M){
			// 我们保证了实线r大于虚线r。
			if(lr[idx][1]<r){
				break;
			}
			trl.add(lr[idx][0],1);
			++idx;
		}
		// 要求实线l小于虚线r大于虚线l
		Ans[ind]=trl.query(l,r);
#ifdef LOCAL
	if(ind==2){
		debug(trl.query(l,r));
		debug(l);
		debug(r);
		FOR(j,1,idx){
			debug(lr[j][0]);
			debug(lr[j][1]);
		}
	}
#endif
	}
	sort(que+1,que+1+Q);
	sort(lr+1,lr+1+M);
	idx=1;
	FOR(i,1,Q){
		ll l=que[i][0],r=que[i][1],ind=que[i][2];
		while(idx<=M){
			// 我们保证了实线l小于虚线l。
			if(lr[idx][0]>l){
				break;
			}
			trr.add(lr[idx][1],1);
			++idx;
		}
		// 要求实线l小于虚线r大于虚线l
		Ans[ind]+=trr.query(l,r);
#ifdef LOCAL
	if(ind==2){
		debug(trr.query(l,r));
		debug(l);
		debug(r);
		for(int j=1;j<=idx;++j){
			debug(lr[j][0]);
			debug(lr[j][1]);
		}
	}
#endif
	}
	FOR(i,1,Q){
		cout<<Ans[i]<<"\n";
	}
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	// cin>>T;
	// while(T--){
		// solve();
	// }
	solve();
	return 0;
}
/*
AC
https://atcoder.jp/contests/abc405/submissions/66033424
*/

心路历程（WA，TLE，MLE……）

P3398 仓鼠找 sugar（倍增法求LCA）

Posted on 2025-05-21 Edited on 2026-02-16

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。

//        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

源代码

// 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……）

https://www.luogu.com.cn/record/217981621

主要是无法处理这种情况。

源代码

// 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;
}
/*

*/

POJ - 2356-Find a multiple

Posted on 2025-05-20 Edited on 2026-02-16

思路讲解

AC代码

源代码

心路历程（WA，TLE，MLE……）

HDU - 6152-Friend-Graph

Posted on 2025-05-20 Edited on 2026-02-16

思路讲解

就是拉姆齐定理的简单应用

N≥6 一定是 Bad Team。≤5，枚举一下就行

if(N<=5){
		FOR(i,1,N)
		FOR(j,i+1,N)
		FOR(k,j+1,N){
			if(G[i][j]+G[i][k]+G[j][k]==0 || G[i][j]+G[i][k]+G[j][k]==3){
				cout<<"Bad Team!\n";
				return;
			}
		}
		cout<<"Great Team!\n";
	}else{
		cout<<"Bad Team!\n";
	}

AC代码

https://vjudge.net/solution/60815052

源代码

// Problem: Friend-Graph
// Contest: Virtual Judge - HDU
// URL: https://vjudge.net/problem/HDU-6152
// Memory Limit: 32 MB
// Time Limit: 5000 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>(1e6)+10,INF=static_cast<ll>(1e18)+3;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;

ll N,M,K,T;
bool G[3010][3010];
/*

*/
inline void solve(){
	FOR(i,0,N+5)
	FOR(j,0,N+5){
		G[i][j]=0;
	}
	cin>>N;
	FOR(i,1,N-1){
		FOR(j,i+1,N){
			ll val;
			cin>>val;
			G[i][j]=val;
			G[j][i]=val;
		}
	}
	if(N<=5){
		FOR(i,1,N)
		FOR(j,i+1,N)
		FOR(k,j+1,N){
			if(G[i][j]+G[i][k]+G[j][k]==0 || G[i][j]+G[i][k]+G[j][k]==3){
				cout<<"Bad Team!\n";
				return;
			}
		}
		cout<<"Great Team!\n";
	}else{
		cout<<"Bad Team!\n";
	}
	
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>T;
	while(T--){
		solve();
	}
	// solve();
	return 0;
}
/*

*/