1137. Bus Routes Time limit: 1.0 secondMemory limit: 64 MB Several bus routes were in the city of Fishburg. None of the routes shared the same section of road, though common stops and intersections were possible. Fishburg old residents stated that it…

Hyperchannels Time limit: 1.0 secondMemory limit: 64 MB The Galaxy Empire consists of N planets. Hyperchannels exist between most of the planets. New Emperor urged to extend hyperchannels network in such a way, that he can move from any planet to any…

1137 做过一样的 怎么又忘了 再一次搜超时 不用回溯 #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> #include<vector> #include<queue> using namespace std; #define N 10010 int n,m; vector<i…

We have a list of bus routes. Each routes[i] is a bus route that the i-th bus repeats forever. For example if routes[0] = [1, 5, 7], this means that the first bus (0-th indexed) travels in the sequence 1->5->7->1->5->7->1->... forever…

We have a list of bus routes. Each routes[i]is a bus route that the i-th bus repeats forever. For example if routes[0] = [1, 5, 7], this means that the first bus (0-th indexed) travels in the sequence 1->5->7->1->5->7->1->... forever.…

题目链接:http://uoj.ac/problem/117 题目大意: 解题思路:先判断度数: 若G为有向图,欧拉回路的点的出度等于入度. 若G为无向图,欧拉回路的点的度数位偶数. 然后判断连通性,并且输出路径需要用套圈法(其实我也不是很懂). 还学了一些骚操作: ①用链式前向星存图,如果是有向图,那idx隔两个存一条边,如果是无向图则idx隔一个存一条边,且idx从2开始.这样写的作用就是在寻无向图路径时可以良好地标记,比如第一条无向边里idx=2.3分别对应一条正反边,2和3除2都对应1,…

We have a list of bus routes. Each routes[i] is a bus route that the i-th bus repeats forever. For example if routes[0] = [1, 5, 7], this means that the first bus (0-th indexed) travels in the sequence 1->5->7->1->5->7->1->... forever…

We have a list of bus routes. Each routes[i] is a bus route that the i-th bus repeats forever. For example if routes[0] = [1, 5, 7], this means that the first bus (0-th indexed) travels in the sequence 1->5->7->1->5->7->1->... forever…

题目如下: We have a list of bus routes. Each routes[i] is a bus route that the i-th bus repeats forever. For example if routes[0] = [1, 5, 7], this means that the first bus (0-th indexed) travels in the sequence 1->5->7->1->5->7->1->... f…

hdu 5552 Bus Routes 考虑有环的图不方便,可以考虑无环连通图的数量,然后用连通图的数量减去就好了. 无环连通图的个数就是树的个数,又 prufer 序我们知道是 $ n^{n-2} $ 其中又由于有 $ n-1 $ 个边,每个边可以涂色,所以总共无环的方案数量是 $ m^{n-1} n^{n-2} $ 那么现在就要算连通图的数量了.这个不如不连通图的数量好算. 不连通图的数量怎么算呢,原本想的是容斥,但是貌似不好实现,看了题解发现一种神仙思路.考虑固定一个点,并且让这个点连出一…