The meaning of problems:
given an undirected graph, then there is \ (q, q \ leq 30 \) group asked asking each have a \ (X \) , the number of points the answer \ ((a, b) \ ) it \ (ab \) minimum cut of no more than \ (the X-\) .
Ideas:
This title approach to minimum cut trees ... this thing is probably to the current point set arbitrarily select two points \ (s, t \) as the source and sink, and then find the current minimum cut, after two sets even the smallest edge cut weight; and two sets recursion process.
Obviously there will be only a last element in the collection, the last form is a tree.
Nature has a minimal cut tree: For trees \ (u, v \) points, which on the right side of the path is the minimum of the minimum of two cut.
This thanks to understand it?
When the final disposition of each time point in accordance with a set of reordering and recursion, another note update \ (ans \) enumeration range, the minimum any pair of points then the two sets of cutting no more than \ (d \) .
See, an intuitive understanding of the minimal cut tree: Portal
code show as below:
/*
* 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;
}