UVa 10766 Organising the Organisation（矩阵树定理）

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UVA 10766 Organising the Organisation

https://vjudge.net/problem/UVA-10766 题意: n个员工,除总经理外每个人只能有一个直接上级有m对人不能成为直接的上下级关系规定k为总经理问员工分级方案无向图生成树的个数与根无关所以直接用矩阵树定理即可 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long LL; ][]; ][]; int m…

UVa 10766 Organising the Organisation（矩阵树定理）

https://vjudge.net/problem/UVA-10766 题意: 给出n, m, k.表示n个点,其中m条边不能直接连通,求生成树个数. 思路: 这也算个裸题,把可以连接的边连接起来,然后矩阵树计算一下即可. #include<iostream> #include<algorithm> #include<cstring> #include<cstdio> #include<sstream> #include<vector&g…

UVa 10766 Organising the Organisation (生成树计数)

题意:给定一个公司的人数,然后还有一个boss,然后再给定一些人,他们不能成为直属上下级关系,问你有多少种安排方式(树). 析:就是一个生成树计数,由于有些人不能成为上下级关系,也就是说他们之间没有边,没说的就是有边,用Matrix-Tree定理,很容易就能得到答案,注意题目给定的可能有重复的. 对于基尔霍夫矩阵,就是度数矩阵,减去邻接矩阵,一处理就OK了. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #i…

Uva 10766 Organising the Organisation (Matrix_tree 生成树计数)

题目描述: 一个由n个部门组成的公司现在需要分层,但是由于员工间的一些小小矛盾,使得他们并不愿意做上下级,问在满足他们要求以后有多少种分层的方案数? 解题思路: 生成树计数模板题,建立Kirchhoff矩阵,利用Matrix_tree定理求解. Kirchhoff矩阵:假设G为n*n矩阵,C为G的入度矩阵(i==j时,C[i][j]等于i的入度;i!=j时,C[i][j]等于零),A为G的邻接矩阵,那么就有Kirchhoff矩阵等于C-A. Matrix_tree定理:G的不同生成树的个数等于其…

生成树的计数(基尔霍夫矩阵)：UVAoj 10766 Organising the Organisation SPOJ HIGH - Highways

HIGH - Highways 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 ci…

[spoj104][Highways] (生成树计数+矩阵树定理+高斯消元)

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…