传送！
这道题比较有思维难度。
首先考虑这么个事：树上有(m)个被标记的点，求使这些点连通的边集的最小总长度。
做法是把这些点按dfs序排序，然后求dfs序相邻的两个点的距离（包括第(m)个点和第1个点）。这些距离之和就是答案的二倍！
知道了这个后，这道题就好办了：对于加入或删除一个点，我们用set维护以dfs序为关键字的序列并及时改变答案即可。
用set的时候记得特判，有些小细节不太好写。
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<queue>
#include<set>
#include<assert.h>
#include<ctime>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
#define forE(i, x, y) for(int i = head[x], y; ~i && (y = e[i].to); i = e[i].nxt)
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e5 + 5;
const int N = 18;
In ll read()
{
ll ans = 0;
char ch = getchar(), las = ' ';
while(!isdigit(ch)) las = ch, ch = getchar();
while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
if(las == '-') ans = -ans;
return ans;
}
In void write(ll x)
{
if(x < 0) x = -x, putchar('-');
if(x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
In void MYFILE()
{
#ifndef mrclr
freopen(".in", "r", stdin);
freopen(".out", "w", stdout);
#endif
}
char op[2];
int n, m;
struct Edge
{
int nxt, to, w;
}e[maxn << 1];
int head[maxn], ecnt = -1;
In void addEdge(int x, int y, int w)
{
e[++ecnt] = (Edge){head[x], y, w};
head[x] = ecnt;
}
ll sum[maxn];
int dfsx[maxn], pos[maxn], cnt = 0;
int dep[maxn], fa[N + 2][maxn];
In void dfs(int now, int _f)
{
dfsx[now] = ++cnt; pos[cnt] = now;
for(int i = 1; (1 << i) <= dep[now]; ++i)
fa[i][now] = fa[i - 1][fa[i - 1][now]];
forE(i, now, v)
{
if(v == _f) continue;
dep[v] = dep[now] + 1, fa[0][v] = now;
sum[v] = sum[now] + e[i].w;
dfs(v, now);
}
}
In int lca(int x, int y)
{
if(dep[x] < dep[y]) swap(x, y);
for(int i = N; i >= 0; --i)
if(dep[fa[i][x]] >= dep[y]) x = fa[i][x];
if(x == y) return x;
for(int i = N; i >= 0; --i)
if(fa[i][x] ^ fa[i][y]) x = fa[i][x], y = fa[i][y];
return fa[0][x];
}
#define sint set<int>::iterator
set<int> s;
ll ans = 0;
In int pre(int x)
{
sint it = s.lower_bound(dfsx[x]);
return it == s.begin() ? pos[*(--s.end())] : pos[*(--it)];
}
In int nxt(int x)
{
sint it = s.lower_bound(dfsx[x]);
++it;
return it == s.end()? pos[*s.begin()] : pos[*it];
}
In ll len(int x, int y)
{
int z = lca(x, y);
ll tp = sum[x] + sum[y] - (sum[z] << 1);
return tp;
}
In void update(int x, bool flg)
{
if(flg) s.insert(dfsx[x]);
int a = pre(x), b = nxt(x);
if(!flg) s.erase(dfsx[x]);
if(!a) return;
ll ab = len(a, b), ax = len(a, x), xb = len(x, b);
if(flg) ans = ans - ab + ax + xb;
else ans = ans - ax - xb + ab;
}
int main()
{
// MYFILE();
Mem(head, -1), ecnt = -1;
n = read();
for(int i = 1; i < n; ++i)
{
int x = read(), y = read(), w = read();
addEdge(x, y, w), addEdge(y, x, w);
}
dep[1] = 1, dfs(1, 0);
m = read();
for(int i = 1; i <= m; ++i)
{
scanf("%s", op);;
if(op[0] == '+') update(read(), 1);
else if(op[0] == '-') update(read(), 0);
else write(ans >> 1), enter;
}
return 0;
}