[色多项式] UOJ #308. 【UNR #2】UOJ拯救计划 & SRM 717 div1 AcyclicOrientation

一个图的 k 染色数是关于 k 的 n 次(?) 多项式
称为色多项式
那么这里模6 我们只要知道模2和模3的值
然后分类讨论下就好了
一张图的0染色数是0，1染色数等于[m=0]，2染色数与二分图的联通块个数有关

#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>#define cl(x) memset(x,0,sizeof(x))using namespace std;inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void read(int &x){  char c=nc(),b=1;  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;  for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}const int N=100005;struct edge{  int u,v,next;}G[N<<2];int head[N],inum;inline void add(int u,int v,int p){  G[p].u=u; G[p].v=v; G[p].next=head[u]; head[u]=p;}#define V G[p].vint fat[N],depth[N];int flag=0;inline void dfs(int u,int fa){  fat[u]=fa; depth[u]=depth[fa]+1;  for (int p=head[u];p;p=G[p].next)    if (V!=fa){      if (!depth[V]){    dfs(V,u);      }else if (depth[V]<depth[u]){    flag|=((depth[u]-depth[V])%2==0);      }    }}int n,m,K;inline int Pow(int a,int b,int P){  int ret=1;  for (;b;b>>=1,a=a*a%P)    if (b&1)      ret=ret*a%P;  return ret;}int main(){  int T,x,y;  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  read(T);  while (T--){    read(n); read(m); read(K); inum=0; cl(head);     for (int i=1;i<=m;i++)      read(x),read(y),add(x,y,++inum),add(y,x,++inum);    int mod2,mod3;    if (~K&1)      mod2=0;    else      mod2=(m==0);    if (K%3==0)      mod3=0;    else if (K%3==1)      mod3=(m==0);    else{      cl(depth); cl(fat); int cnt=0; flag=0;      for (int i=1;i<=n;i++)    if (!depth[i])      cnt++,dfs(i,0);      if (flag) mod3=0;      else mod3=Pow(2,cnt,3);    }    for (int i=0;i<6;i++)      if (i%2==mod2 && i%3==mod3)    printf("%d\n",i);  }  return 0;}

TC那个题是求把无向图定向成DAG的方案数
有定理，也可见这里

The number of Acyclic Orientations of an undirected graph is given by ( − 1)N⋅P(−1), where P is the chromatic polynomial of the given graph.

// BEGIN CUT HERE  #include<conio.h>#include<sstream>// END CUT HERE  #include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<iostream>#include<algorithm>#include<vector>#include<map>#include<string>#include<set>#define cl(x) memset(x,0,sizeof(x))using namespace std;typedef long long ll;const int N=105;const int M=N*N;struct edge{  int u,v,next;}G[M];int head[N],inum;inline void add(int u,int v,int p){  G[p].u=u; G[p].v=v; G[p].next=head[u]; head[u]=p;}#define V G[p].vint depth[N];int flag=0;inline void dfs(int u,int fa){  depth[u]=depth[fa]+1;  for (int p=head[u];p;p=G[p].next)    if (!depth[V])      dfs(V,u);    else if (depth[V]<depth[u] && (depth[u]-depth[V])%2==0)      flag=1;}inline int Pow(int a,int b){  int ret=1;  for (;b;b>>=1,a=a*a%3)    if (b&1)      ret=ret*a%3;  return ret;}class AcyclicOrientation{public:  int count(int n, vector <int> u, vector <int> v){    cl(head); inum=0; cl(depth); flag=0;    for (int i=0;i<(int)u.size();i++)      add(u[i]+1,v[i]+1,++inum),add(v[i]+1,u[i]+1,++inum);    int m2=(u.size()==0),m3,C=0;    for (int i=1;i<=n;i++)      if (!depth[i])    dfs(i,0),C++;    if (flag)      m3=0;    else      m3=Pow(2,n+C);    for (int i=0;i<6;i++)      if (i%2==m2 && i%3==m3)    return i;  }  // BEGIN CUT HEREpublic:  void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); }private:  template <typename T> string print_array(const vector<T> &_V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = _V.begin(); iter != _V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }  void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }  void test_case_0() { int Arg0 = 3; int Arr1[] = {0, 1, 0}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arr2[] = {2, 2, 1}; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[0]))); int Arg3 = 0; verify_case(0, Arg3, count(Arg0, Arg1, Arg2)); }  void test_case_1() { int Arg0 = 5; int Arr1[] = {0, 1, 2, 3}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arr2[] = {1, 2, 3, 4}; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[0]))); int Arg3 = 4; verify_case(1, Arg3, count(Arg0, Arg1, Arg2)); }  void test_case_2() { int Arg0 = 4; int Arr1[] = {0, 3}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arr2[] = {1, 1}; vector <int> Arg2(Arr2, Arr2 + (sizeof(Arr2) / sizeof(Arr2[0]))); int Arg3 = 4; verify_case(2, Arg3, count(Arg0, Arg1, Arg2)); }  // END CUT HERE};// BEGIN CUT HEREint main(){  AcyclicOrientation ___test;  ___test.run_test(-1);  getch() ;  return 0;}// END CUT HERE