【Leetcode】882. Reachable Nodes In Subdivided Graph 882. 细分图中的可到达结点

解法

我个人觉得就是一种BFS
优先遍历距离近的点，所以不用队列要用优先队列
最开始是一边遍历一边加，而且要考虑到从i往j用的点数，还有从j往i用的点数，以及它们重合了怎么办

class Solution(object):
    def reachableNodes(self, edges, M, N):
        """
        :type edges: List[List[int]]
        :type M: int
        :type N: int
        :rtype: int
        """
        from collections import defaultdict
        from heapq import heappush,heappop
        adj = defaultdict(list)
        for i,j,k in edges:
            adj[i].append((j,k))
            adj[j].append((i,k))
        bfs = [(0,0)]
        ans = 0
        used = defaultdict(lambda:defaultdict(int))
        mark = set()
        while bfs:
            d,i = heappop(bfs)
            # print i,d
            if i in mark:continue
            mark.add(i)
            ans += 1
            for j,k in adj[i]:
                able = k-used[j][i]
                if k>0 and not able:continue
                used[i][j] = min(able,M-d)
                # print i,j,able,used[i][j]
                ans += used[i][j]
                if d+k+1<=M and j not in mark:
                    heappush(bfs,(d+k+1,j))
        return ans

看了答案之后，发现不需要边遍历边加，对于一条边，我们可以算出来从i到j会用a点，从j到i会用b点
在a+b和k之间取最小值就可以了

class Solution(object):
    def reachableNodes(self, edges, M, N):
        """
        :type edges: List[List[int]]
        :type M: int
        :type N: int
        :rtype: int
        """
        from collections import defaultdict
        from heapq import heappush,heappop
        adj = defaultdict(list)
        for i,j,k in edges:
            adj[i].append((j,k))
            adj[j].append((i,k))
        bfs = [(0,0)]
        used = defaultdict(lambda:defaultdict(int))
        mark = set()
        while bfs:
            d,i = heappop(bfs)
            if d>M:break
            # print i,d
            if i in mark:continue
            mark.add(i)
            for j,k in adj[i]:
                used[i][j] = min(k,M-d)
                if d+k+1<=M and j not in mark:
                    heappush(bfs,(d+k+1,j))
        ans = len(mark)
        for i,j,k in edges:
            ans += min(k,used[i][j]+used[j][i])
        return ans