CSU 1805 three capitals (Matrix tree theorem +best theorem)

http://acm.csu.edu.cn/csuoj/problemset/problem?pid=1805

Test instructions

A and B have a side, a and G have a B-bar, and B and G have a C-side. Now go through all the edges from point a, and then go back to point a, and ask how many ways there are.

Ideas:

16 Hunan Province, the title of the problem is to find the number of Euro-loop, and the count of spanning tree has a certain connection.

First of all, the magic of the best theorem, which is what the devil theorem, anyway, nothing to find out about the theorem of the article ...

$ec (g) =t_s (g) \cdot deg (s)! \cdot \prod_{v\in v,\ v\ne s} (deg (v)-1)!,\ t_s (G): =$ number of outward trees with s root.

The number of outward trees in the graph is almost the same as the number of spanning trees of the graph, in short, The Matrix tree theorem, but a little bit different place.

The structure of the Kirchhoff matrix is not quite the same, after all, one is an no-map, and one is a graph of the direction:

The degree matrix in the no-map is the degree of each vertex, and the degree matrix in the graph is the degree of each vertex.

The adjacency matrix is the number of sides of the u->v, which is similar to the non-graph.

Then there is the calculation of the matrix:

Number of spanning trees for an n-1 graph = values of arbitrary-order master

Number of outward trees of the graph = the value of the principal of the order in which the root node is removed

Note that this topic also need to calculate the number of combinations, for example, A and b there is a side, then I can choose the x edge for the edge, then the remaining a-x edge for the out side, in this process will have a different choice method. The total is $c (a,x) *c (b,y) *c (c,z) $.

1#include <iostream>2#include <algorithm>3#include <cstring>4#include <cstdio>5#include <sstream>6#include <vector>7#include <stack>8#include <queue>9#include <cmath>Ten#include <map> One#include <Set> A using namespacestd; -typedefLong Longll; -typedef pair<int,ll>PLL; the Const intINF =0x3f3f3f3f; - Const intmaxn= -+5; - Constll mod=1e9+7; -  + inta,b,c; -ll f[100005]; +ll a[5][5]; A  at voidInit () - { -f[0]=1; -      for(intI=1; i<=100005; i++) f[i]=f[i-1]*i%MoD; - } -  in ll Qpow (ll a,ll N) - { toll ans=1; +      while(n) -     { the         if(n&1) ans=ans*a%MoD; *a=a*a%MoD; $n>>=1;Panax Notoginseng     } -     returnans; the } +  A ll C (ll n,ll m) the { +     return(F[n]*qpow (f[m],mod-2)%mod) *qpow (f[n-m],mod-2)%MoD; - } $  $ ll Calc () - { -     return(a[1][1]*a[2][2]%mod-a[1][2]*a[2][1]%MOD+MOD)%MoD; the } - Wuyi intMain () the { -     //freopen ("In.txt", "R", stdin); Wu init (); -      while(~SCANF ("%d%d%d",&a,&b,&c)) About     { $         if((A+C) &1|| (a+b) &1|| (B+C) &1) {puts ("0");Continue;} -ll ans=0; -          for(intI=0; i<=a;i++)//enumerates the number of edges of a a->b -         { Amemset (A,0,sizeof(A)); +a[0][0]= (A+B)/2; thea[1][1]= (A+C)/2; -a[2][2]= (B+C)/2; $a[0][1]=-i; thea[1][0]=-(A-i); thea[0][2]=-(a[0][0]-i); thea[2][0]=-(b+a[0][2]); thea[1][2]=-(a[1][1]+a[1][0]); -a[2][1]=-(c+a[1][2]); in             if(a[0][2]>0|| a[2][0]>0|| a[1][2]>0|| a[2][1]>0)Continue; the  thell res= (C (a,i) *c (c,-a[1][2])%mod) *c (b,-a[0][2])%MoD; About  theRes= (Res*calc ())%MoD; the              for(intI=1;i<3; i++) res=res*f[a[i][i]-1]%MoD; theres=res*f[a[0][0]]%MoD; +ans= (ans+res)%MoD; -         } theprintf"%lld\n", ans);Bayi     } the     return 0; the}

CSU 1805 three capitals (Matrix tree theorem +best theorem)