大意: 给定一个基环树, 求删除一条环上的边使得直径最小.
直径分两种情况
- 环上点延伸的树内的直径
- 两个环上点的树内深度最大的点匹配
第一种情况直接树形dp求一下, 第二种情况枚举删除的环边, 线段树维护一下即可.
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head const int N = 1e6+10;
int n,dep[N],vis[N];
struct _ {int to,w;} fa[N];
vector<_> g[N];
int a[N], v[N], top;
ll mx[N], b[N], sum[N], ans, ans2;
struct node {
ll m1,m2,v;
node operator + (const node &rhs) const {
node ret;
ret.m1=max(m1,rhs.m1);
ret.m2=max(m2,rhs.m2);
ret.v=max(v,rhs.v);
ret.v=max(ret.v,m2+rhs.m1);
return ret;
}
} tr[N<<2]; void get(int x, int y) {
if (dep[x]<dep[y]) return;
v[++top] = y;
a[top+1] = fa[y].w;
for (; x!=y; x=fa[x].to) v[++top] = x, a[top+1] = fa[x].w;
}
void dfs(int x, int f) {
dep[x] = dep[f]+1;
for (_ e:g[x]) if (e.to!=f) {
int y = e.to;
fa[y] = {x,e.w};
if (dep[y]) get(x,y);
else dfs(y,x);
}
}
void dfs2(int x) {
vis[x] = 1;
for (_ e:g[x]) if (!vis[e.to]) {
int y = e.to;
dfs2(y);
ans = max(ans, mx[x]+mx[y]+e.w);
mx[x] = max(mx[x], mx[y]+e.w);
}
} void build(int o, int l, int r) {
if (l==r) tr[o].m1=b[l]+sum[l],tr[o].m2=b[l]-sum[l],tr[o].v=0;
else {
build(ls),build(rs);
tr[o]=tr[lc]+tr[rc];
}
}
node qry(int o, int l, int r, int ql, int qr) {
if (ql<=l&&r<=qr) return tr[o];
if (mid>=qr) return qry(ls,ql,qr);
if (mid<ql) return qry(rs,ql,qr);
return qry(ls,ql,qr)+qry(rs,ql,qr);
} int main() {
scanf("%d", &n);
REP(i,1,n) {
int u, v, w;
scanf("%d%d%d",&u,&v,&w);
g[u].pb({v,w});
g[v].pb({u,w});
}
dfs(1,0);
REP(i,1,top) vis[v[i]]=1;
REP(i,1,top) dfs2(v[i]),b[i]=mx[v[i]];
REP(i,top+1,2*top-1) b[i] = b[i-top];
REP(i,top+2,2*top-1) a[i] = a[i-top];
REP(i,1,2*top-1) sum[i] = sum[i-1]+a[i];
build(1,1,2*top-1);
ans2 = 1e18;
REP(i,1,top) ans2 = min(ans2, qry(1,1,2*top-1,i,i+top-1).v);
printf("%lld\n",max(ans,ans2));
}