今天玩魔灵玩多了,耽误了时间,回去宿舍又没电. /*********************************************/ Bus Fair Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 900    Accepted Submission(s): 444 Problem Description You are no…

Bus Fair Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 600    Accepted Submission(s): 293 Problem Description You are now in Foolish Land. Once moving in Foolish Land you found that there is…

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…

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…

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…

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…

题意 给定 \(n\) 个点,任意连边,每条边有 \(m\) 种颜色可选,求带环连通图的方案数. \(1\leq n\leq 10000\) \(1\leq m < 2^{31}\) 思路 直接求带环连通图显然比较难求,正难则反,考虑容斥.用连通图的个数减去无环连通图(树)的个数. \(n\) 个节点的无根树,每个节点有区别,可以直接套用公式 \(n^{n-2}\) .而再考虑边的颜色,就是 \(m^{n-1}n^{n-2}\) . 我们设 \(n\) 个点,考虑边的颜色,构成不同连通图的方案数…

题意 给一个图,每个点的出度为1,每个点的权值为1或者2.给Q个询问,问是否能找到一条路径的权值和M. 思路 由于每个点的出度为1,所以必然存在环.又因为c[i]只能取1或者2,可以组成任意值,所以只要有c[i] == 1 就可以造成任何数.没1时能得到任意偶数 代码 [cpp] #include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <s…