欲学矩阵树定理必先自宫学习一些行列式的姿势 然后做一道例题 #include <iostream> #include <cstring> #include <cstdio> #include <cmath> using namespace std; typedef long long ll; int T, n, m, uu, vv; double w[15][15]; const double eps=1e-7; void gauss(){ n--; for…

裸题 注意: 1.消元时判断系数为0,退出 2.最后乘ans要用double.... #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; typedef long long ll; ; ; inline int read(){ ; '){c=getchar();} +c…

SPOJ104 Highways Description In some countries building highways takes a lot of time- Maybe that's because there are many possiblities to construct a network of highways and engineers can't make up their minds which one to choose. Suppose we have a l…

In some countries building highways takes a lot of time... Maybe that's because there are many possiblities to construct a network of highways and engineers can't make up their minds which one to choose. Suppose we have a list of cities that can be c…

https://blog.csdn.net/zhaoruixiang1111/article/details/79185927 为了学一个矩阵树定理 从行列式开始学(就当提前学线代了.. 论文生成树的计数及其应用 矩阵数定理: 截图来自于上述论文 裸题. #include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> using namespace std; ; ; double…

[SPOJ]Highways(矩阵树定理) 题面 Vjudge 洛谷 题解 矩阵树定理模板题 无向图的矩阵树定理: 对于一条边\((u,v)\),给邻接矩阵上\(G[u][v],G[v][u]\)加一 对于一条边\((u,v)\),给度数矩阵上\(D[u][u],D[v][v]\)加一 定义霍尔基夫矩阵\(C=D-G\) 将基尔霍夫矩阵去除任意一行和任意一列之后, 得到一个\((n-1)*(n-1)\)的行列式\(C\) 求解这个行列式的值,最后的\(|det(C)|\)就是结果 #includ…

题目: In some countries building highways takes a lot of time... Maybe that's because there are many possiblities to construct a network of highways and engineers can't make up their minds which one to choose. Suppose we have a list of cities that can…

传送门 题意: 给定一个一边点数为n,另一边点数为m,共有n*m条边的带标号完全二分图$K_{n,m}$ 求生成树个数 1 <= n,m,p <= 10^18 显然不能暴力上矩阵树定理 看过推到完全图的生成树个数后这道题也不难做 构建出基尔霍夫矩阵,找一个主子式,所有行加起来放一行上,用这一行消消消就发现最后对角线上有$n-1$个$m$和$m-1$个$n$和$1$个$1$ 然后要用快速乘...蒟蒻第一次用快速乘... #include <iostream> using namesp…

4596: [Shoi2016]黑暗前的幻想乡 Time Limit: 20 Sec  Memory Limit: 256 MBSubmit: 559  Solved: 325[Submit][Status][Discuss] Description 四年一度的幻想乡大选开始了,最近幻想乡最大的问题是很多来历不明的妖 怪涌入了幻想乡,扰乱了幻想乡昔日的秩序.但是幻想乡的建制派妖怪(人类) 博丽灵梦和八云紫等人整日高谈所有妖怪平等,幻想乡多元化等等,对于幻想乡 目前面临的种种大问题却给不出合适的解…

[LOJ#6072]苹果树(矩阵树定理,折半搜索,容斥) 题面 LOJ 题解 emmmm,这题似乎猫讲过一次... 显然先\(meet-in-the-middle\)搜索一下对于每个有用的苹果数量,满足权值小于\(lim\)的方案数 ,那么只需要考虑它们构成生成树的方案数就好了. 显然有用的可以和所有的有用的或者是坏的连边,好的但不有用的只能和坏的连边,而坏的随意. 但是这样子算出来的结果是至多,因此还需要额外容斥一下计算生成树的个数. #include<iostream> #include&…