Znzryb's Blog

P3398 仓鼠找 sugar（倍增法求LCA）

Posted on 2025-05-21 Edited on 2026-04-06

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 2026-04-06

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 2026-04-06

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;
}
/*

*/

心路历程（WA，TLE，MLE……）

HDU - 5917-Instability

Posted on 2025-05-19 Edited on 2026-04-06

思路讲解

根据 Ramsey定理，大于等于6个的集合，肯定存在一个子集的边都是红色或者都是蓝色，即为团还是为孤立点；

即6 个人中至少存在3人相互认识或者相互不认识。

FOR(i,6,N){		// 根据Ramsey定理的结论
	ans+=C(N,i);
	ans%=mod;
}

然后3-5个点的枚举就可以了

AC代码

https://vjudge.net/solution/60814540

源代码

// Problem: Instability
// Contest: Virtual Judge - HDU
// URL: https://vjudge.net/problem/HDU-5917
// Memory Limit: 64 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,0,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,oT;
bool G[55][55];
/*

*/
ll frac[60],infrac[60];
inline ll C(ll n,ll k){
	ll res=frac[n]*infrac[n-k];
	res%=mod;
	res*=infrac[k];
	res%=mod;
	return res;
}
inline ll binpow(ll a,ll k){
	ll res=1;
	while(k){
		if(k&1) res*=a;
		a*=a;
		a%=mod;
		res%=mod;
		k>>=1;
	}
	return res;
}
inline void solve(){
	cin>>N>>M;
	FOR(i,0,N+3){
		FOR(j,0,N+3){
			G[i][j]=false;
		}
	}
	
	FOR(i,1,M){
		ll u,v;
		cin>>u>>v;
		G[u][v]=true;
		G[v][u]=true;
	}
	ll ans=0;
	FOR(i,6,N){		// 根据Ramsey定理的结论
		ans+=C(N,i);
		ans%=mod;
	}
	FOR(i,1,N){
		FOR(j,i+1,N){
			FOR(k,j+1,N){
				bool alTrue=false;
				// if(G[i][j]==false && G[i][k]==false && G[k][j]==false){
					// ans+=1;
					// alTrue=true;
				// }else if(G[i][j] && G[i][k] && G[j][k]){
					// ans+=1;
					// alTrue=true;
				// }
				auto cal=[&](ll i,ll j,ll k){
					if(G[i][j]==false && G[i][k]==false && G[k][j]==false){
						alTrue=true;
					}
					if(G[i][j] && G[i][k] && G[j][k]){
						alTrue=true;
					}
				};
				cal(i,j,k);
				if(alTrue){
					ans+=1;
				}
				FOR(i4,k+1,N){
					alTrue=false;
					cal(i,j,k);
					cal(i,j,i4);
					cal(i,i4,k);
					cal(i4,j,k);
					if(alTrue){
						ans+=1;
					}
					FOR(i5,i4+1,N){
						alTrue=false;
						cal(i,j,k);
						cal(i,j,i4);
						cal(i,i4,k);
						cal(i4,j,k);
						cal(i,j,i5);
						cal(i,i5,k);
						cal(i5,j,k);
						cal(i5,j,i4);
						cal(i5,i4,k);
						cal(i5,i,i4);
						if(alTrue){
							ans+=1;
						}
					}
				}
			}
		}
	}
	ans%=mod;
	if(ans<0) ans+=mod;
	ll Case=oT-T;
	cout<<"Case #"<<Case<<": "<< ans <<"\n";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>T;
	oT=T;
	frac[0]=1;
	FOR(i,1,50){
		frac[i]=frac[i-1]*i;
		frac[i]%=mod;
	}
	infrac[0]=1;
	FOR(i,1,50){
		infrac[i]=binpow(frac[i],mod-2);
	}
#ifdef LOCAL
    debugN(frac,50);
    cerr<<"\n";
    debugN(infrac,50);
    debug(C(4,2));
#endif
	while(T--){
		solve();
	}
	// solve();
	return 0;
}
/*
AC https://vjudge.net/solution/60814540
1
7 8
1 2
2 3
3 4
4 5
5 6
6 7
1 7
1 4
ans:Case #1: 53

1
5 6
1 2
2 3
3 4
2 5
4 5
1 5
ans: Case #1: 4
*/

心路历程（WA，TLE，MLE……）

小心 infrac[0]=0;

infrac[0]=1;
FOR(i,1,50){
	infrac[i]=binpow(frac[i],mod-2);
}

HDU - 1205-吃糖果

Posted on 2025-05-19 Edited on 2026-04-06

思路讲解

隔板法

如果说有maxA个挡板，至少要有maxA-1个数才能让隔板不相邻。

AC代码

https://vjudge.net/solution/60783771

源代码

// Problem: 吃糖果
// Contest: Virtual Judge - HDU
// URL: https://vjudge.net/problem/HDU-1205
// Memory Limit: 32 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> 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,A[MAXN];
/*

*/
inline void solve(){
	cin>>N;
	ll sum=0,maxA=0;
	FOR(i,1,N){
		cin>>A[i];
		sum+=A[i];
		maxA=max(A[i],maxA);
	}
	// 如果说有maxA个挡板，至少要有maxA-1个数才能让隔板不相邻
	if(sum-maxA<maxA-1){
		cout<<"No\n";
	}else{
		cout<<"Yes\n";
	}
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>T;
	while(T--){
		solve();
	}
	// solve();
	return 0;
}
/*
AC
https://vjudge.net/solution/60783771
*/