题意：
给出一个无向图，之后有\(q,q\leq 30\)组询问，每组询问有一个\(x\)，回答有多少点对\((a,b)\)其\(a-b\)最小割不超过\(x\)。
思路：
这个题做法要最小割树...这个东西大概就是对于当前点集任意选择两个点\(s,t\)作为源点和汇点，然后求出当前最小割，之后两个集合连边为最小割权值；然后两个集合递归下去处理。
显然最后集合中只会存在一个元素，那么最后形成的就是一颗树。
最小割树有一个性质：对于树上\(u,v\)两点，其路径上的边权最小值即为两点的最小割。
这个感谢理解一下？
最后处理的时候每次将点按照集合重排序然后递归下去，另外注意一下更新\(ans\)时的枚举范围，这时两个集合中的任意一对点的最小割都不会超过\(d\)。
直观理解最小割树参见：传送门
代码如下：
/*
* Author: heyuhhh
* Created Time: 2019/10/31 21:20:48
*/
#include <bits/stdc++.h>
#define MP make_pair
#define fi first
#define se second
#define sz(x) (int)(x).size()
#define all(x) (x).begin(), (x).end()
#define INF 0x3f3f3f3f
#define Local
#ifdef Local
#define dbg(args...) do { cout << #args << " -> "; err(args); } while (0)
void err() { std::cout << '\n'; }
template<typename T, typename...Args>
void err(T a, Args...args) { std::cout << a << ' '; err(args...); }
#else
#define dbg(...)
#endif
void pt() {std::cout << '\n'; }
template<typename T, typename...Args>
void pt(T a, Args...args) {std::cout << a << ' '; pt(args...); }
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
//head
const int N = 155, M = 10005;
#define _S heyuhhh
template <class T>
struct Dinic{
struct Edge{
int v, next;
T flow, w;
Edge(){}
Edge(int v, int next, T flow, T w) : v(v), next(next), flow(flow), w(w) {}
}e[M << 1];
int head[N], tot;
int dep[N];
bool vis[N];
void init() {
memset(head, -1, sizeof(head)); tot = 0;
}
void adde(int u, int v, T w, T rw = 0) {
e[tot] = Edge(v, head[u], w, w);
head[u] = tot++;
e[tot] = Edge(u, head[v], rw, rw);
head[v] = tot++;
}
bool BFS(int _S, int _T) {
memset(dep, 0, sizeof(dep));
queue <int> q; q.push(_S); dep[_S] = 1;
while(!q.empty()) {
int u = q.front(); q.pop();
for(int i = head[u]; ~i; i = e[i].next) {
int v = e[i].v;
if(!dep[v] && e[i].flow > 0) {
dep[v] = dep[u] + 1;
q.push(v);
}
}
}
return dep[_T] != 0;
}
T dfs(int _S, int _T, T a) {
T flow = 0, f;
if(_S == _T || a == 0) return a;
for(int i = head[_S]; ~i; i = e[i].next) {
int v = e[i].v;
if(dep[v] != dep[_S] + 1) continue;
f = dfs(v, _T, min(a, e[i].flow));
if(f) {
e[i].flow -= f;
e[i ^ 1].flow += f;
flow += f;
a -= f;
if(a == 0) break;
}
}
if(!flow) dep[_S] = -1;
return flow;
}
T dinic(int _S, int _T) {
T max_flow = 0;
while(BFS(_S, _T)) max_flow += dfs(_S, _T, INF);
return max_flow;
}
void color(int u) {
vis[u] = true;
for(int i = head[u]; i != -1; i = e[i].next) {
int v = e[i].v;
if(vis[v] == false && e[i].flow) color(v);
}
}
void pre() {
memset(vis, 0, sizeof(vis));
for(int i = 0; i < tot; i++) e[i].flow = e[i].w;
}
};
Dinic <int> solver;
int n, m;
int c[N][N];
int a[N], ans[N][N];
int tmp1[N], tmp2[N];
void solve(int l, int r) {
if(l >= r) return;
solver.pre();
int S = a[l], T = a[r];
int d = solver.dinic(S, T);
solver.color(S);
for(int i = 1; i <= n; i++) if(solver.vis[i]) {
for(int j = 1; j <= n; j++) if(!solver.vis[j]) {
ans[i][j] = ans[j][i] = min(ans[i][j], d);
}
}
int t1 = 0, t2 = 0;
for(int i = l; i <= r; i++) {
if(solver.vis[a[i]]) tmp1[++t1] = a[i];
else tmp2[++t2] = a[i];
}
for(int i = l; i <= l + t1 - 1; i++) a[i] = tmp1[i - l + 1];
for(int i = l + t1; i <= r; i++) a[i] = tmp2[i - t1 - l + 1];
solve(l, l + t1 - 1); solve(l + t1, r);
}
void run(){
cin >> n >> m;
solver.init();
memset(c, 0, sizeof(c));
memset(ans, INF, sizeof(ans));
for(int i = 1; i <= m; i++) {
int u, v, w; cin >> u >> v >> w;
c[u][v] += w;
c[v][u] = c[u][v];
}
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
if(c[i][j]) solver.adde(i, j, c[i][j]);
}
}
for(int i = 1; i <= n; i++) a[i] = i;
solve(1, n);
int q; cin >> q;
for(int k = 1; k <= q; k++) {
int x; cin >> x;
int res = 0;
for(int i = 1; i <= n; i++)
for(int j = i + 1; j <= n; j++)
if(ans[i][j] <= x) ++res;
cout << res << '\n';
}
cout << '\n';
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cout << fixed << setprecision(20);
int T; cin >> T;
while(T--) run();
return 0;
}