假设我们有一个无向加权连通图。该图有 n 个节点,它们的标签从 1 到 n。从开始到结束的路径是一系列节点,例如 [z0, z1, z2, ..., zk] 这里 z0 是开始节点,zk 是结束节点,在 zi 和 zi+1 之间有一条边,其中 0 <=我 <= k-1。路径的距离是路径边缘的权重值之和。Letdist(x)表示距节点n和节点x的最短距离。受限路径是一条特殊路径,它也满足dist(zi)> dist(zi+1) where 0 <= i <= k-1。因此,我们必须找到从节点 1 到节点 n 的受限路径的数量。如果答案太大,则返回模 10^9 + 7 的答案。
所以,如果输入像
那么输出将为 3,因为存在三个受限路径 (1,2,5),(1,2,3,5),(1,3,5)。
让我们看看以下实现以更好地理解 -
from collections import defaultdict from heapq import heappop, heappush def solve(n, edges): graph = defaultdict(dict) for u, v, w in edges: graph[u][v] = w graph[v][u] = w paths = [0] * (n+1) paths[n] = 1 dists = [-1] * (n+1) q = [(0, n)] while q: dist, node = heappop(q) if dists[node] != -1: continue dists[node] = dist for v, w in graph[node].items(): if dists[v] == -1: heappush(q, (dist + w, v)) elif dists[v] < dists[node]: paths[node] += paths[v] if node == 1: return paths[node] % (10**9 + 7) return 0 n = 5 edges = [(1,2,3),(1,3,3),(2,3,1),(1,4,2),(5,2,2),(3,5,1),(5,4,10)] print(solve(n, edges))
5,[(1,2,3),(1,3,3),(2,3,1),(1,4,2),(5,2,2),(3,5,1),(5,4,10)]输出结果
3