leetcode(33)图系列题目

1615. 最大网络秩

直接模拟即可

class Solution:
    def maximalNetworkRank(self, n: int, roads: List[List[int]]) -> int:
        adjs = defaultdict(set)  # 注意是集合
        for u, v in roads:
            adjs[u].add(v)
            adjs[v].add(u)
        res = 0
        for i in range(n):
            for j in range(i + 1, n):
                cur  = len(adjs[i]) + len(adjs[j])  # 注意是当前节点边数之和
                if j in adjs[i]:
                    cur -= 1
                res = max(res, cur)
        return res

Dijkstra 算法相关题目

882. 细分图中的可到达节点

heappush(q,(newd, y))比heappush(q,(y, newd))效率快一些，可能因为默认为小根堆，大的newd会直接放在下面，不用排序

class Solution:
    def reachableNodes(self, edges: List[List[int]], maxMoves: int, n: int) -> int:
        g = [[] for _ in range(n)]
        for u, v, cnt in edges:
            g[u].append((v, cnt + 1))
            g[v].append((u, cnt + 1))
        
        dist = self.dijkstra(g, 0)
        # 原始图可到达的节点数
        res = sum(d <= maxMoves for d in dist)
        for u, v, cnt in edges:
            # 到达原始图中的节点后剩余的步数
            a = max(0, maxMoves - dist[u])
            b = max(0, maxMoves - dist[v])
            # 每条边中可到达的新节点数
            res += min(a + b, cnt)
        return res 

    def dijkstra(self, g, start):
        dist = [inf] * len(g)
        dist[start] = 0
        q = [(0, start)]
        while q:
            d, x = heappop(q)
            if dist[x] < d:  # 已经是最短路径，如果=说明有一条新的最短路，还是要进行下面的操作
                continue
            for y, w in g[x]:
                newd = dist[x] + w 
                if newd < dist[y]:
                    dist[y] = newd 
                    heappush(q,(newd, y))
        return dist