# https://atcoder.jp/contests/practice2/submissions/20654283 import heapq class mcf_graph_int_cost: """It solves Minimum-cost flow problem. """ def __init__(self, n): """It creates a directed graph with n vertices and 0 edges. Cap and Cost are the type of the capacity and the cost, respectively. Constraints ----------- > 0 <= n <= 10 ** 8 > Cap and Cost are int. Complexity ---------- > O(n) """ self.n = n self._edges = [] def add_edge(self, from_, to, cap, cost): """It adds an edge oriented from `from_` to `to` with capacity `cap` and cost `cost`. It returns an integer k such that this is the k-th edge that is added. Constraints ----------- > 0 <= from_, to < n > 0 <= cap, cost Complexity ---------- > O(1) amortized """ # assert 0 <= from_ < self.n # assert 0 <= to < self.n # assert 0 <= cap # assert 0 <= cost m = len(self._edges) self._edges.append(self.__class__.edge(from_, to, cap, 0, cost)) return m class edge: def __init__(self, from_, to, cap, flow, cost): self.from_ = from_ self.to = to self.cap = cap self.flow = flow self.cost = cost def get_edge(self, i): """It returns the current internal state of the edges. The edges are ordered in the same order as added by add_edge. Constraints ----------- > 0 <= i < m Complexity ---------- > O(1) """ # assert 0 <= i < len(self._edges) return self._edges[i] def edges(self): """It returns the current internal state of the edges. The edges are ordered in the same order as added by add_edge. Complexity ---------- > O(m), where m is the number of added edges. """ return self._edges.copy() def _dual_ref(self, s, t): self.dist = [2147483647] * self.n self.vis = [False] * self.n self.que_min.clear() self.que.clear() que_push_que = [] self.dist[s] = 0 self.que_min.append(s) while self.que_min or self.que or que_push_que: if self.que_min: v = self.que_min.pop() else: while que_push_que: heapq.heappush(self.que, que_push_que.pop()) v = heapq.heappop(self.que) & 4294967295 if self.vis[v]: continue self.vis[v] = True if v == t: break dual_v = self.dual[v] dist_v = self.dist[v] for i in range(self.start[v], self.start[v + 1]): e = self.elist[i] if not e.cap: continue cost = e.cost - self.dual[e.to] + dual_v if self.dist[e.to] - dist_v > cost: dist_to = dist_v + cost self.dist[e.to] = dist_to self.prev_e[e.to] = e.rev if dist_to == dist_v: self.que_min.append(e.to) else: que_push_que.append((dist_to << 32) + e.to) if not self.vis[t]: return False for v in range(self.n): if not self.vis[v]: continue self.dual[v] -= self.dist[t] - self.dist[v] return True def _csr(self): m = len(self._edges) self.edge_idx = [0] * m redge_idx = [0] * m degree = [0] * self.n edges = [] for i, e in enumerate(self._edges): self.edge_idx[i] = degree[e.from_] degree[e.from_] += 1 redge_idx[i] = degree[e.to] degree[e.to] += 1 edges.append((e.from_, self.__class__._edge( e.to, -1, e.cap - e.flow, e.cost))) edges.append((e.to, self.__class__._edge( e.from_, -1, e.flow, -e.cost))) self.start = [0] * (self.n + 1) self.elist = [0] * len(edges) for v, w in edges: self.start[v + 1] += 1 for i in range(1, self.n + 1): self.start[i] += self.start[i-1] counter = self.start.copy() for v, w in edges: self.elist[counter[v]] = w counter[v] += 1 for i, e in enumerate(self._edges): self.edge_idx[i] += self.start[e.from_] redge_idx[i] += self.start[e.to] self.elist[self.edge_idx[i]].rev = redge_idx[i] self.elist[redge_idx[i]].rev = self.edge_idx[i] def slope(self, s, t, flow_limit=2147483647): """Let g be a function such that g(x) is the cost of the minimum cost s-t flow when the amount of the flow is exactly x. g is known to be piecewise linear. It returns g as the list of the changepoints, that satisfies the followings. > The first element of the list is (0, 0). > Both of element[0] and element[1] are strictly increasing. > No three changepoints are on the same line. > (1) The last element of the list is (x, g(x)), where x is the maximum amount of the s-t flow. > (2) The last element of the list is (y, g(y)), where y = min(x, flow_limit). Constraints ----------- Let x be the maximum cost among all edges. > s != t > You can't call min_cost_slope or min_cost_max_flow multiple times. > 0 <= nx <= 2 * 10 ** 9 + 1000 Complexity ---------- > O(F (n + m) log (n + m)), where F is the amount of the flow and m is the number of added edges. """ # assert 0 <= s < self.n # assert 0 <= t < self.n # assert s != t self._csr() self.dual = [0] * self.n self.dist = [2147483647] * self.n self.prev_e = [0] * self.n self.vis = [False] * self.n flow = 0 cost = 0 prev_cost_per_flow = -1 result = [(0, 0)] self.que = [] self.que_min = [] while flow < flow_limit: if not self._dual_ref(s, t): break c = flow_limit - flow v = t while v != s: c = min(c, self.elist[self.elist[self.prev_e[v]].rev].cap) v = self.elist[self.prev_e[v]].to v = t while v != s: e = self.elist[self.prev_e[v]] e.cap += c self.elist[e.rev].cap -= c v = self.elist[self.prev_e[v]].to d = -self.dual[s] flow += c cost += c * d if prev_cost_per_flow == d: result.pop() result.append((flow, cost)) prev_cost_per_flow = d for i in range(len(self._edges)): e = self.elist[self.edge_idx[i]] self._edges[i].flow = self._edges[i].cap - e.cap return result def flow(self, s, t, flow_limit=2147483647): """It augments the flow from s to t as much as possible. It returns the amount of the flow and the cost. (1) It augments the s-t flow as much as possible. (2) It augments the s-t flow as much as possible, until reaching the amount of flow_limit. Constraints ----------- > same as mcf_graph.slope Complexity ---------- > same as mcf_graph.slope """ return self.slope(s, t, flow_limit)[-1] class _edge: def __init__(self, to, rev, cap, cost): self.to = to self.rev = rev self.cap = cap self.cost = cost from collections import Counter k,n,m = map(int,input().split()) a = Counter(map(int,input().split())) b = list(map(int,input().split())) g = mcf_graph_int_cost(n+2) for i,v in a.items(): g.add_edge(0,i,v,0) for i,v in enumerate(b,1): if v: g.add_edge(i,n+1,v,0) for _ in range(m): u,v,d = map(int,input().split()) g.add_edge(u,v,k,d); g.add_edge(v,u,k,d) print(g.flow(0,n+1,k)[1])