题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1690 Bus System Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6569    Accepted Submission(s): 1692 Problem Description Because of the huge popula…

Problem Description Because of the huge population of China, public transportation is very important. Bus is an important transportation method in traditional public transportation system. And it's still playing an important role even now. The bus sy…

Bus System Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6082    Accepted Submission(s): 1560 Problem Description Because of the huge population of China, public transportation is very importa…

题目大意:给出若干巴士不同价格的票的乘坐距离范围,现在有N个站点,有M次询问,查询任意两个站点的最小花费 解析:由于是多次查询不同站点的最小花费,所以用弗洛伊德求解 时间复杂度(O^3) 比较基础的弗洛伊德 #include <iostream> #include <cstdio> #include <cstring> using namespace std; #define INF 1000000000000 typedef __int64 LL; ; __int64…

Bus System Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5190    Accepted Submission(s): 1275 [题目链接]http://acm.hdu.edu.cn/showproblem.php?pid=1690 [解题思路]SPFA求最短路径问题,将每个站之间的距离转化为相应的价格从而建立两站点相连的…

Problem Description Because of the huge population of China, public transportation is very important. Bus is an important transportation method in traditional public transportation system. And it’s still playing an important role even now.The bus sys…

Problem Description Because of the huge population of China, public transportation is very important. Bus is an important transportation method in traditional public transportation system. And it’s still playing an important role even now. The bus sy…

Bus System Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8055    Accepted Submission(s): 2121 Problem Description Because of the huge population of China, public transportation is very importa…

Bus Pass Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 667    Accepted Submission(s): 271 Problem Description You travel a lot by bus and the costs of all the seperate tickets are starting to…

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…