[ZJOI2011] minimal cut (minimal cut tree template)

https://www.luogu.com.cn/problem/P3329

Minimum cut tree is useful not only do the bare title, about this theorem, a class of problems will have a deeper thinking.

Minimal cut tree:
each taking two points u, v, find their cutting, and cutting the tree at the minimum edge connected to them, a weight of the cut.
Then follow can come S T, and can come, into two points, recursively achievements.

Minimum of two points in the original cut, that is the minimum value of the right side of the tree.

Easy to show that the minimal cut <= minimum edge weight of the tree, but to prove just that more difficult, bloggers not going to.

Code：

---

#include<bits/stdc++.h>
#define fo(i, x, y) for(int i = x, _b = y; i <= _b; i ++)
#define ff(i, x, y) for(int i = x, _b = y; i <  _b; i ++)
#define fd(i, x, y) for(int i = x, _b = y; i >= _b; i --)
#define ll long long
#define pp printf
#define hh pp("\n")
using namespace std;

const int N = 6005;

int n, m;
int x, y, z;

int fi[N], to[N], nt[N], r[N], tot = 1;

void link(int x, int y, int z) {
	nt[++ tot] = fi[x], to[tot] = y, r[tot] = z, fi[x] = tot;
	nt[++ tot] = fi[y], to[tot] = x, r[tot] = z, fi[y] = tot;
}


int dis[N], d[N], d0, S, T;

int bfs() {
	fo(i, 0, n) dis[i] = 1e9;
	dis[S] = 0, d[d0 = 1] = S;
	for(int i = 1; i <= d0; i ++) {
		int x = d[i];
		for(int j = fi[x]; j; j = nt[j]) if(r[j]) {
			int y = to[j];
			if(dis[y] == 1e9) {
				dis[y] = dis[x] + 1;
				d[++ d0] = y;
			}
		}
	}
	return dis[T] != 1e9;
}

int dg(int x, int flow) {
	if(x == T) return flow;
	int use = 0;
	for(int i = fi[x]; i; i = nt[i]) if(r[i] && dis[x] + 1 == dis[to[i]]) {
		int t = dg(to[i], min(flow - use, r[i]));
		use += t, r[i] -= t, r[i ^ 1] += t;
		if(use == flow) return use;
	}
	return use;
}

int mark[N];

void dfs(int x) {
	if(mark[x]) return;
	mark[x] = 1;
	for(int i = fi[x]; i; i = nt[i]) if(r[i])
		dfs(to[i]);
}

#define V vector<int>
#define pb push_back
#define re resize
#define si size()

int ans[N][N];

void cl() {
	for(int i = 2; i <= tot; i += 2) {
		int s = r[i] + r[i ^ 1];
		r[i] = r[i ^ 1] = s / 2;
	}
}

void dg(V a) {
	if(a.si <= 1) return;
	cl();
	S = a[0], T = a[1];
	int sum = 0;
	while(bfs()) sum += dg(S, 1 << 30);
	fo(i, 1, n) mark[i] = 0;
	dfs(S);
	fo(i, 1, n) if(mark[i])
		fo(j, 1, n) if(!mark[j]) {
			if(sum < ans[i][j])
				ans[i][j] = ans[j][i] = sum;
		}
	V b, c; b.clear(); c.clear();
	ff(i, 0, a.si) if(mark[a[i]])
		b.pb(a[i]); else c.pb(a[i]);
	dg(b); dg(c);
}

void cl_e() {
	fo(i, 1, n) fi[i] = 0;
	tot = 1;
}

int main() {
	int T;
	scanf("%d", &T);
	fo(iT, 1, T) {
		cl_e();
		scanf("%d %d", &n, &m);
		fo(i, 1, m) {
			scanf("%d %d %d", &x, &y, &z);
			link(x, y, z);
		}
		fo(i, 1, n) fo(j, 1, n) ans[i][j] = 1e9;
		V a; a.re(n);
		fo(i, 1, n) a[i - 1] = i;
		dg(a);
		int q, x;
		scanf("%d", &q);
		fo(ii, 1, q) {
			scanf("%d", &x);
			int s = 0;
			fo(i, 1, n) fo(j, i + 1, n)
				s += ans[i][j] <= x;
			pp("%d\n", s);
		}
		hh;
	}
}