## Solución al problema número POTHOLE de Spoj - POTHOLE.

Cualquier duda no dudes en contactar.

``````// Adjacency list implementation of FIFO push relabel maximum flow
// with the gap relabeling heuristic.  This implementation is
// significantly faster than straight Ford-Fulkerson.  It solves
// random problems with 10000 vertices and 1000000 edges in a few
// seconds, though it is possible to construct test cases that
// achieve the worst-case.
//
// Running time:
//     O(|V|^3)
//
// INPUT:
//     - graph, constructed using AddEdge()
//     - source
//     - sink
//
// OUTPUT:
//     - maximum flow value
//     - To obtain the actual flow values, look at all edges with
//       capacity > 0 (zero capacity edges are residual edges).

#include <cmath>
#include <vector>
#include <iostream>
#include <queue>

using namespace std;

typedef long long LL;

const int INF = 1000000000;

struct Edge {
int from, to, cap, flow, index;
Edge(int from, int to, int cap, int flow, int index) :
from(from), to(to), cap(cap), flow(flow), index(index) {}
};

struct PushRelabel {
int N;
vector<vector<Edge> > G;
vector<LL> excess;
vector<int> dist, active, count;
queue<int> Q;

PushRelabel(int N) : N(N), G(N), excess(N), dist(N), active(N), count(2*N) {}

void AddEdge(int from, int to, int cap) {
G[from].push_back(Edge(from, to, cap, 0, G[to].size()));
if (from == to) G[from].back().index++;
G[to].push_back(Edge(to, from, 0, 0, G[from].size() - 1));
}

void Enqueue(int v) {
if (!active[v] && excess[v] > 0) { active[v] = true; Q.push(v); }
}

void Push(Edge &e) {
int amt = int(min(excess[e.from], LL(e.cap - e.flow)));
if (dist[e.from] <= dist[e.to] || amt == 0) return;
e.flow += amt;
G[e.to][e.index].flow -= amt;
excess[e.to] += amt;
excess[e.from] -= amt;
Enqueue(e.to);
}

void Gap(int k) {
for (int v = 0; v < N; v++) {
if (dist[v] < k) continue;
count[dist[v]]--;
dist[v] = max(dist[v], N+1);
count[dist[v]]++;
Enqueue(v);
}
}

void Relabel(int v) {
count[dist[v]]--;
dist[v] = 2*N;
for (int i = 0; i < G[v].size(); i++)
if (G[v][i].cap - G[v][i].flow > 0)
dist[v] = min(dist[v], dist[G[v][i].to] + 1);
count[dist[v]]++;
Enqueue(v);
}

void Discharge(int v) {
for (int i = 0; excess[v] > 0 && i < G[v].size(); i++) Push(G[v][i]);
if (excess[v] > 0) {
if (count[dist[v]] == 1)
Gap(dist[v]);
else
Relabel(v);
}
}

LL GetMaxFlow(int s, int t) {
count[0] = N-1;
count[N] = 1;
dist[s] = N;
active[s] = active[t] = true;
for (int i = 0; i < G[s].size(); i++) {
excess[s] += G[s][i].cap;
Push(G[s][i]);
}

while (!Q.empty()) {
int v = Q.front();
Q.pop();
active[v] = false;
Discharge(v);
}

LL totflow = 0;
for (int i = 0; i < G[s].size(); i++) totflow += G[s][i].flow;
}
};

int main() {
int t;
cin >> t;
for (int i = 0; i < t; i++) {
int n;
cin >> n;
PushRelabel pr(n);
for (int j = 0; j < n-1; j++) {
int m;
cin >> m;
for (int k = 0; k < m; k++) {
int p;
cin >> p;
p--;
int cap = (j == 0 || p == n-1) ? 1 : INF;