18/03: Problem C

18-03-24/c.py

@@ -0,0 +1,50 @@
+## Récupération des entrées
+n, m = [int(i) for i in input().split()]
+
+# adj = [list() for i in range(n+1)]
+anti_adj = [list() for i in range(n+1)]
+for i in range(m):
+    u, v = [int(i) for i in input().split()]
+    anti_adj[v].append(u)
+    # adj[u].append(v)
+
+k = int(input())
+p = [int(i) for i in input().split()]
+
+## Initialisation des structures
+dist = [-1 for i in range(n+1)]
+succ = {i+1:set() for i in range(n)}
+size = {i+1:0 for i in range(n)}
+
+non_vus = set((i+1 for i in range(n) if i+1 != p[-1]))
+queue = [p[-1]]
+dist[p[-1]] = 0
+
+## Calcul des prochains avec leur coût, BFS par la fin
+while len(queue) > 0:
+    nextQ = []
+    while len(queue) > 0:
+        s = queue.pop()
+        for pred in anti_adj[s]:
+            if pred in non_vus:
+                non_vus.remove(pred)
+                nextQ.append(pred)
+                dist[pred] = dist[s]+1
+            if dist[pred] == dist[s]+1:
+                succ[pred].add(s)
+                size[pred] += 1
+    queue = nextQ
+
+## Calcul final
+mini, maxi = 0, 0
+position = p[0]
+for i in range(1, k):
+    next_pos = p[i]
+    if next_pos not in succ[position]:
+        mini += 1
+        maxi += 1
+    elif len(succ[position]) > 1:
+        maxi += 1
+    position = next_pos
+
+print(mini, maxi)