我们对于一棵树，我们找一个根，然后统计过这个点的路径有多少符合要求。怎么搞呢？我们可以先维护一个数据结构，然后把先把根作为一个距离自己为0的点放进去，然后对于每一棵子树，先找出所有的与之前的数据结构的东西进行统计，然后放入数据结构，递归每一棵子树，就可以搞了。为了保证复杂度，所以每一次选重心提起来。还是比较好些，然后我写的splay就莫名的慢，想快的可以写treap之类的(' '     ) 

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 using namespace std;
  6 
  7 typedef long long ll;
  8 const ll maxn = 100010;
  9 const ll INF = 0x3f3f3f3f;
 10 
 11 struct node {
 12     ll data, weight, size; 
 13     node *son[2], *fa;
 14 }e[maxn]; ll ne = 0;
 15 node* root;
 16 
 17 void update(node* x) {
 18     x-> size = x-> weight;
 19     for(ll i = 0; i < 2; ++ i) {
 20         if(x-> son[i]) x-> size += x-> son[i]-> size;
 21     }
 22 }
 23 
 24 void rotate(node *x, ll f) {
 25     node* t = x-> fa;
 26     if(t-> fa) {
 27         if(t-> fa-> son[0] == t) t-> fa-> son[0] = x;
 28         else t-> fa-> son[1] = t;
 29     }
 30     x-> fa = t-> fa, x-> size = t-> size, t-> fa = x;
 31     t-> son[f] = x-> son[!f];
 32     if(x-> son[!f]) x-> son[!f]-> fa = t;
 33     x-> son[!f] = t;
 34     update(t);
 35 }
 36 
 37 void splay(node* x, node* f) {
 38     while(x-> fa != f) {
 39         if(x-> fa-> fa == f) {
 40         ll a = x-> fa-> son[0] == x ? 0 : 1;
 41         rotate(x, a);
 42         }
 43         else {
 44             node* t = x-> fa, *r = t-> fa;
 45             ll a = r-> son[0] == t ? 0 : 1;
 46             ll b = t-> son[0] == x ? 0 : 1;
 47             if(a == b) rotate(t, a), rotate(x, b);
 48             else rotate(x, b), rotate(x, a);
 49         }
 50     }
 51     if(x-> fa == NULL) root = x;
 52 }
 53 
 54 node* newnode(ll v) {
 55     node* x = e + ne ++;
 56     x-> data = v, x-> size = x-> weight = 1;
 57     x-> son[0] = x-> son[1] = NULL;
 58     return x;
 59 }
 60 
 61 void insert(ll v) {
 62     node* x = root;
 63     while(x) {
 64         x-> size ++;
 65         if(x-> data == v) {
 66             x-> weight ++;
 67             splay(x, NULL); break;
 68         }
 69         else {
 70             ll a = x-> data > v ? 0 : 1;
 71             if(x-> son[a]) x = x-> son[a];
 72             else {
 73                 x-> son[a] = newnode(v);
 74                 x-> son[a]-> fa = x;
 75                 splay(x-> son[a], NULL);
 76                 break;
 77             }
 78         }
 79     }
 80     if(!x) root = newnode(v), root-> fa = NULL;
 81 }
 82 
 83 ll work(ll v) {
 84     node *x = root; ll ret = 0;
 85     while(x) {
 86         if(x-> data <= v) {
 87             ret += x-> weight + (x-> son[0] ? x-> son[0]-> size : 0);
 88             x = x-> son[1];
 89         }
 90         else x = x-> son[0];
 91     }
 92     return ret;
 93 }
 94 
 95 struct edge {
 96     ll t, d;
 97     edge* next;
 98 }se[maxn * 2], *head[maxn]; ll oe = 0;
 99 ll n, m, a[maxn], top = 0;
100 ll s[maxn], stop = 0;
101 
102 void addedge(ll f, ll t, ll d) {
103     se[oe].t = t, se[oe].d = d, se[oe].next = head[f], head[f] = se + oe ++;
104 }
105 
106 bool vis[maxn]; ll size[maxn], dis[maxn];
107 
108 void dfs(ll x, ll fa) {
109     size[x] = 1; s[++ stop] = x;
110     for(edge* p = head[x]; p; p = p-> next) if(p-> t != fa && !vis[p-> t]) dfs(p-> t, x), size[x] += size[p-> t];
111 }
112 
113 void find_dis(ll x, ll fa) {
114     a[++ top] = x;
115     for(edge* p = head[x]; p; p = p-> next) {
116         if(!vis[p-> t] && p-> t != fa) dis[p-> t] = dis[x] + p-> d, find_dis(p-> t, x);
117     }
118 }
119 
120 ll int_get() {
121     ll x = 0; char c = (char)getchar(); bool f = 0;
122     while(!isdigit(c) && c != '-') c = (char)getchar();
123     if(c == '-') c = (char)getchar(), f = 1;
124     while(isdigit(c)) {
125         x = x * 10 + (int)(c - '0');
126         c = (char)getchar();
127     }
128     if(f) x = -x;
129     return x;
130 }
131 
132 void read() {
133     n = int_get();
134     for(ll i = 1; i <= n - 1; ++ i) {
135         ll u, v, w; u = int_get(), v = int_get(), w = int_get();
136         addedge(u, v, w), addedge(v, u, w);
137     }
138     m = int_get();
139 }
140 
141 ll ans = 0;
142 
143 void sov(ll x) {
144     stop = 0;
145     dfs(x, 0); ll tot = size[x];
146     if(tot == 1) return; 
147     ll Max = INF, pos = x; 
148     for(ll j = 1; j <= stop; ++ j) {
149         ll i = s[j];
150         ll tmax = 0;
151         for(edge* p = head[i]; p; p = p-> next) if(!vis[p-> t] && size[x] > size[p-> t]) tmax = max(tmax, size[p-> t]);
152         tmax = max(tot - size[i], tmax);
153         if(Max > tmax) Max = tmax, pos = i;
154     }
155     ll rt = pos; root = NULL; ne = 0;
156     insert(0);
157     for(edge* p = head[rt]; p; p = p-> next) {
158         if(!vis[p-> t]) {
159             top = 0, dis[p-> t] = p-> d, find_dis(p-> t, rt);
160             for(ll i = 1; i <= top; ++ i) ans += work(m - dis[a[i]]);
161             for(ll i = 1; i <= top; ++ i) insert(dis[a[i]]);
162         }
163     }
164     vis[rt] = 1;
165     for(edge* p = head[rt]; p; p = p-> next) {
166         if(!vis[p-> t]) sov(p-> t);
167     }
168 }
169 
170 int main() {
171     freopen("test.in", "r", stdin);
172     freopen("test.out", "w", stdout);
173     read(); 
174     memset(vis, 0, sizeof(vis));
175     sov(1);
176     printf("%lld\n", ans);
177     return 0;
178 }
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