先建出来点分树，以每个点为根开线段树，维护点分子树内编号为$[l,r]$的儿子到根的距离最小值
每次查询$x$开始，沿着点分树向上跑，在每个点的线段树的$[l,r]$区间里都查一遍取$min$即可
因为题目让我们求最小值，所以出现重复经过同一条路径的情况并不会让答案变坏
如果让我们求最大值呢？似乎可以用主席树搞搞？(仅仅是口胡有空再想)
1 #include <cmath> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #define N1 100005 6 #define M1 505 7 #define ll long long 8 #define uint unsigned int 9 #define ush unsigned short 10 using namespace std; 11 const int inf=0x3f3f3f3f; 12 13 template <typename _T> void read(_T &ret) 14 { 15 ret=0; _T fh=1; char c=getchar(); 16 while(c<'0'||c>'9'){ if(c=='-') fh=-1; c=getchar(); } 17 while(c>='0'&&c<='9'){ ret=ret*10+c-'0'; c=getchar(); } 18 ret=ret*fh; 19 } 20 21 struct Edge{ 22 int to[N1*2],nxt[N1*2],val[N1*2],head[N1],cte; 23 void ae(int u,int v,int w) 24 { cte++; to[cte]=v; nxt[cte]=head[u]; val[cte]=w; head[u]=cte; } 25 }e; 26 27 int lg[N1*2],de; 28 namespace Tree{ 29 int dis[N1],dep[N1],fa[N1],st[N1],ff[N1*2][19],cur; //ord[N1*2], 30 void dfs1(int x) 31 { 32 int j,v; st[x]=++cur; ff[cur][0]=x; 33 for(j=e.head[x];j;j=e.nxt[j]) 34 { 35 v=e.to[j]; if(v==fa[x]) continue; 36 fa[v]=x; dis[v]=dis[x]+e.val[j]; dep[v]=dep[x]+1; 37 dfs1(v); ff[++cur][0]=x; 38 } 39 } 40 void init() 41 { 42 dep[1]=1; dfs1(1); 43 int i,j; 44 for(i=2,lg[1]=0;i<=cur;i++) lg[i]=lg[i>>1]+1; 45 for(j=1;j<=lg[cur];j++) 46 for(i=1;i+(1<<j)-1<=cur;i++) 47 ff[i][j]=dep[ff[i][j-1]] < dep[ff[i+(1<<(j-1))][j-1]] ? ff[i][j-1] : ff[i+(1<<(j-1))][j-1]; 48 } 49 inline int Dis(int x,int y) 50 { 51 int i=st[x],j=st[y],F; if(i>j) swap(i,j); 52 F=dep[ff[i][lg[j-i+1]]] < dep[ff[j-(1<<lg[j-i+1])+1][lg[j-i+1]]] ? ff[i][lg[j-i+1]] : ff[j-(1<<lg[j-i+1])+1][lg[j-i+1]]; 53 return dis[x]+dis[y]-2*dis[F]; 54 } 55 }; 56 57 struct SEG{ 58 struct node{ int ls,rs,mi; }; 59 node a[N1*230]; int root[N1],tot; 60 inline void pushup(int rt) 61 { 62 a[rt].mi=inf; 63 if(a[rt].ls) a[rt].mi=min(a[rt].mi,a[a[rt].ls].mi); 64 if(a[rt].rs) a[rt].mi=min(a[rt].mi,a[a[rt].rs].mi); 65 } 66 void upd(int x,int l,int r,int &rt,int w) 67 { 68 if(!rt) rt=++tot; 69 if(l==r){ a[rt].mi=w; return; } 70 int mid=(l+r)>>1; 71 if(x<=mid) upd(x,l,mid,a[rt].ls,w); 72 else upd(x,mid+1,r,a[rt].rs,w); 73 pushup(rt); 74 } 75 int qmi(int L,int R,int l,int r,int rt) 76 { 77 if(!rt) return inf; 78 if(L<=l&&r<=R) return a[rt].mi; 79 int mid=(l+r)>>1; int ans=inf; 80 if(L<=mid) ans=min(ans,qmi(L,R,l,mid,a[rt].ls)); 81 if(R>mid) ans=min(ans,qmi(L,R,mid+1,r,a[rt].rs)); 82 return ans; 83 } 84 }s; 85 86 using Tree::Dis; 87 88 int n; 89 int sz[N1],ms[N1],use[N1],pfa[N1],SZ,G; 90 91 void gra(int x,int dad,int now) 92 { 93 int j,v; sz[x]=1; ms[x]=0; 94 if(now) s.upd(x,1,n,s.root[now],Dis(x,now)); 95 for(j=e.head[x];j;j=e.nxt[j]) 96 { 97 v=e.to[j]; if(v==dad || use[v]) continue; 98 gra(v,x,now); 99 sz[x]+=sz[v]; ms[x]=max(ms[x],sz[v]); 100 } 101 ms[x]=max(ms[x],SZ-sz[x]); 102 if(ms[x]<ms[G]) G=x; 103 } 104 105 void main_dfs(int x) 106 { 107 int j,v; use[x]=1; 108 s.upd(x,1,n,s.root[x],0); 109 for(j=e.head[x];j;j=e.nxt[j]) 110 { 111 v=e.to[j]; if(use[v]) continue; 112 SZ=sz[v]; G=0; gra(v,x,x); pfa[G]=x; 113 main_dfs(G); 114 } 115 } 116 117 int main() 118 { 119 scanf("%d",&n); 120 int i,j,x,y,w,Q,q,l,r,D,ans; 121 for(i=1;i<n;i++) 122 { 123 read(x), read(y), read(w); 124 e.ae(x,y,w); e.ae(y,x,w); 125 } 126 Tree::init(); 127 ms[0]=n; SZ=n; gra(1,0,0); main_dfs(G); 128 scanf("%d",&Q); 129 for(q=1;q<=Q;q++) 130 { 131 read(l), read(r), read(x); ans=inf; 132 for(y=x;y;y=pfa[y]) 133 { 134 D=Dis(x,y); 135 ans=min(ans,D+s.qmi(l,r,1,n,s.root[y])); 136 } 137 printf("%d ",ans); 138 } 139 return 0; 140 }