题意：给定一张图，保证 $1$ 号点到其他所有点的最短路径是唯一的，求：对于点 $i$，不经过 $1$ 到 $i$ 最短路径上最后一条边的最短路.
题解：可以先建出最短路树，然后枚举每一条非树边.
对于一条非树边，影响的只是 $(u,lca)$，$(lca,v)$ 这些点的答案，然后你发现可以写成 $dep[a]-dep[u]+val[i]+dep[v]$
对于 $a$ 来说，第一项是固定的，后面的几项对于一个链来说都是相同的，所以直接用 $LCT$ 维护区间最小值就行了.
#include <bits/stdc++.h>
#define N 100003
#define M 200005
#define ll long long
#define inf 1000000000000
#define inf2 10000000000
#define setIO(s) freopen(s".in","r",stdin) , freopen(s".out","w",stdout)
using namespace std;
ll output[N];
vector<int>G[N];
int lst[N],from[N],n,m;
namespace Dij
{
ll val[M<<1],d[N];
int hd[N],to[M<<1],nex[M<<1],done[N],edges,s;
struct Node
{
int u;
ll dis;
Node(int u=0,ll dis=0):u(u),dis(dis){}
bool operator<(Node b) const { return b.dis<dis; }
};
priority_queue<Node>q;
void addedge(int u,int v,ll c)
{
nex[++edges]=hd[u],hd[u]=edges,to[edges]=v,val[edges]=c;
}
void dijkstra()
{
memset(d,0x3f,sizeof(d));
d[s]=0ll;
q.push(Node(s,0ll));
while(!q.empty())
{
Node e=q.top(); q.pop();
int u=e.u;
if(done[u]) continue;
done[u]=1;
for(int i=hd[u];i;i=nex[i])
{
int v=to[i];
if(d[u]+val[i]<d[v])
{
lst[v]=i;
from[v]=u;
d[v]=d[u]+val[i];
q.push(Node(v, d[v]));
}
}
}
for(int i=2;i<=n;++i) G[from[i]].push_back(i);
}
}
namespace tree
{
int tim;
int size[N],dfn[N],top[N],dep[N],son[N],fa[N];
void dfs1(int u,int ff)
{
fa[u]=ff,size[u]=1,dep[u]=dep[ff]+1;
for(int i=0;i<G[u].size();++i)
{
dfs1(G[u][i],u);
size[u]+=size[G[u][i]];
if(size[G[u][i]]>size[son[u]]) son[u]=G[u][i];
}
}
void dfs2(int u,int tp)
{
top[u]=tp;
if(son[u]) dfs2(son[u], tp);
for(int i=0;i<G[u].size();++i)
if(G[u][i]!=son[u]) dfs2(G[u][i], G[u][i]);
}
int LCA(int x,int y)
{
while(top[x]!=top[y])
{
dep[top[x]]>dep[top[y]]?x=fa[top[x]]:y=fa[top[y]];
}
return dep[x]<dep[y]?x:y;
}
};
struct Link_Cut_Tree
{
#define lson p[x].ch[0]
#define rson p[x].ch[1]
int sta[N];
struct Node
{
int ch[2],tag,f,rev;
ll minn,pu,val;
}p[N];
inline int get(int x) { return p[p[x].f].ch[1]==x; }
inline int isrt(int x) { return !(p[p[x].f].ch[0]==x||p[p[x].f].ch[1]==x); }
inline void pushup(int x)
{
p[x].minn=p[x].val;
if(lson) p[x].minn=min(p[x].minn, p[lson].minn);
if(rson) p[x].minn=min(p[x].minn, p[rson].minn);
}
void markrev(int x)
{
if(x) swap(lson,rson), p[x].rev^=1;
}
void marktag(int x,ll v)
{
if(x)
{
p[x].val=min(p[x].val,v);
if(!p[x].tag || p[x].pu>v) p[x].pu=v, p[x].tag=1;
pushup(x);
}
}
void pushdown(int x)
{
if(!x) return;
if(p[x].tag)
{
if(lson) marktag(lson, p[x].pu);
if(rson) marktag(rson, p[x].pu);
p[x].tag=0;
}
if(p[x].rev)
{
if(lson) markrev(lson);
if(rson) markrev(rson);
p[x].rev=0;
}
}
inline void rotate(int x)
{
int old=p[x].f, fold=p[old].f, which=get(x);
if(!isrt(old)) p[fold].ch[p[fold].ch[1]==old]=x;
p[old].ch[which]=p[x].ch[which^1], p[p[old].ch[which]].f=old;
p[x].ch[which^1]=old, p[old].f=x, p[x].f=fold;
pushup(old), pushup(x);
}
inline void splay(int x)
{
int u=x,v=0,fa;
for(sta[++v]=u; !isrt(u); u=p[u].f) sta[++v]=p[u].f;
for(;v;--v) pushdown(sta[v]);
for(u=p[u].f;(fa=p[x].f)!=u;rotate(x))
if(p[fa].f!=u)
rotate(get(fa)==get(x)?fa:x);
}
void Access(int x)
{
for(int y=0;x;y=x,x=p[x].f)
splay(x), rson=y, pushup(x);
}
void makeroot(int x)
{
Access(x), splay(x), markrev(x);
}
void split(int x,int y)
{
makeroot(x), Access(y), splay(y);
}
int find(int x)
{
for(pushdown(x); rson ; pushdown(x))
{
x=rson;
}
return x;
}
#undef lson
#undef rson
}lct;
void build_tree(int u,int ff)
{
lct.p[u].f=ff;
lct.p[u].val=inf;
lct.pushup(u);
for(int i=0;i<G[u].size();++i) build_tree(G[u][i], u);
}
int main()
{
// setIO("input");
int i,j;
scanf("%d%d",&n,&m);
for(i=1;i<=m;++i)
{
int u,v,c;
scanf("%d%d%d",&u,&v,&c);
Dij::addedge(u,v,1ll*c);
Dij::addedge(v,u,1ll*c);
}
Dij::s=1;
Dij::dijkstra();
tree::dfs1(1,0);
tree::dfs2(1,1);
build_tree(1,0);
for(i=1;i<=n;++i)
{
for(j=Dij::hd[i];j;j=Dij::nex[j])
{
int v=Dij::to[j];
if(lst[v]==j) continue;
int lca=tree::LCA(i,v);
lct.split(v,lca);
int pp=lct.find(lct.p[lca].ch[0]);
if(pp)
{
lct.split(v, pp);
lct.marktag(pp, Dij::d[v]+Dij::d[i]+Dij::val[j]);
}
}
}
for(i=2;i<=n;++i)
{
lct.Access(i);
lct.splay(i);
if(lct.p[i].val>=inf2) printf("-1
");
else printf("%lld
",lct.p[i].val-Dij::d[i]);
}
return 0;
}