import sys input=sys.stdin.readline # 最小費用流 from heapq import heappush, heappop class M1inCostFlow: INF = 10**18 def __init__(self, N): self.N = N self.G = [[] for i in range(N)] def add_edge(self, fr, to, cap, cost): forward = [to, cap, cost, None] backward = forward[3] = [fr, 0, -cost, forward] self.G[fr].append(forward) self.G[to].append(backward) def flow(self, s, t, f): N = self.N; G = self.G INF = MinCostFlow.INF res = 0 H = [0]*N prv_v = [0]*N prv_e = [None]*N d0 = [INF]*N dist = [INF]*N while f: dist[:] = d0 dist[s] = 0 que = [(0, s)] while que: c, v = heappop(que) if dist[v] < c: continue r0 = dist[v] + H[v] for e in G[v]: w, cap, cost, _ = e if cap > 0 and r0 + cost - H[w] < dist[w]: dist[w] = r = r0 + cost - H[w] prv_v[w] = v; prv_e[w] = e heappush(que, (r, w)) if dist[t] == INF: return res return None for i in range(N): H[i] += dist[i] d = f; v = t while v != s: d = min(d, prv_e[v][1]) v = prv_v[v] f -= d res += d * H[t] v = t while v != s: e = prv_e[v] e[1] -= d e[3][1] += d v = prv_v[v] return res import heapq class MinCostFlow: def __init__(self, n, neg=False): self.n=n self.g=[[] for _ in range(n)] self.h=[0]*n self.neg=neg def add_edge(self, frm, to, cap, cost): g=self.g g[frm].append([to, cap, cost, len(g[to])]) g[to].append([frm, 0, -cost, len(g[frm])-1]) def bellman_ford(self, s, t, prevv, preve): n=self.n g=self.g h=self.h INF=10**18 for x in range(n): h[x]=INF h[s]=0 for _ in range(n-1): for x in range(n): for i, e in enumerate(g[x]): if e[1]==0: continue to, cost=e[0], e[2] if h[to]>h[x]+cost: h[to]=h[x]+cost prevv[to]=x preve[to]=i def flow(self, s, t, f): res=0 n=self.n g=self.g h=self.h INF=10**18 LOG=13 #n<2**LOG prevv=[0]*n preve=[0]*n while f>0: if self.neg: self.neg=False self.bellman_ford(s, t, prevv, preve) if h[t]==INF: return -1 else: que=[] dist=[INF]*n dist[s]=0 heapq.heappush(que, s) while que: p=heapq.heappop(que) v=(p&((1<>LOG): continue for i, e in enumerate(g[v]): if e[1]==0: continue to, cost=e[0], e[2] if dist[to]>dist[v]+cost+h[v]-h[to]: dist[to]=dist[v]+cost+h[v]-h[to] prevv[to]=v preve[to]=i heapq.heappush(que, (dist[to]<n+1 ary=[n+1]*5 ary[4]=n d={'y':0,'u':1,'k':2,'i':3} for i in range(n-1,-1,-1): k=d[s[i]] if ary[k]