[矩阵树定理 容斥 meet in middle] Topcoder SRM 551 DIV1 Hard. SweetFruits

枚举最后的树中有多少个是truly sweet的
答案就是

$$\sum_{i=0}^n f_i\times g_i$$

其中，$f_i$ 表示选出 $i$ 个水果使其价值和不超过Limit的方案数
$g_i$ 是有 $i$ 个truly sweet的生成树个数

$f_i$ 可以meet in middle
设图中前 $i$ 个点是truly sweet， $i+1$ 到 $k$ 是sweet的
truly sweet之间两两连边，不甜的水果两两连边，不甜的水果和甜的水果连边

跑矩阵树定理可以得到truly sweet数量之多为 $i$ 的方案数

容斥一下就得到了 $g_i$

// BEGIN CUT HERE  

// END CUT HERE  
#include <vector>  
#include <list>  
#include <map>  
#include <set>  
#include <deque>  
#include <stack>  
#include <bitset>  
#include <algorithm>  
#include <functional>  
#include <numeric>  
#include <utility>  
#include <sstream>  
#include <iostream>  
#include <iomanip>  
#include <cstdio>  
#include <cmath>  
#include <cstdlib>  
#include <ctime>  
#include <queue>
#include <assert.h>
#include <cstring>  
#define pb push_back

using namespace std;  

const int N=105,P=1e9+7;

int n,lim,a[N],f[N],g[N];

vector<int> s,A[N],B[N];

void dfs(int x,int n,int num,int v,vector<int> C[]){
  if(v>lim) return ;
  if(x>=n) return C[num].pb(v);
  dfs(x+1,n,num+1,v+s[x],C);
  dfs(x+1,n,num,v,C);
}

int cnt;

inline void part1(){
  for(int i=0;i<=n;i++) A[i].clear(),B[i].clear();
  dfs(0,cnt/2,0,0,A);
  dfs(cnt/2,cnt,0,0,B);
  for(int i=0;i<=n;i++) if(A[i].size()) sort(A[i].begin(),A[i].end());
  for(int i=0;i<=n;i++) if(B[i].size()) sort(B[i].begin(),B[i].end());
  for(int i=0;i<=n;i++){
    if(!A[i].size()) continue;
    for(int j=0;j<=n;j++){
      if(!B[j].size()) continue;
      int p=0;
      for(int k=B[j].size()-1;~k;k--){
    while(p<A[i].size() && A[i][p]+B[j][k]<=lim) p++;
    (f[i+j]+=p)%=P;
      }
    }
  }
}

typedef int Mat[N][N];

int inv[N],fac[N];

inline int Pow(int x,int y){
  int ret=1;
  for(;y;y>>=1,x=1LL*x*x%P) if(y&1) ret=1LL*ret*x%P;
  return ret;
}

inline int det(Mat a){
  int sp=0;
  for(int i=1;i<n;i++){
    int p; for(p=i;!a[p][i] && p<n;p++);
    if(p==n) return 0;
    if(p!=i){
      for(int j=1;j<n;j++) swap(a[i][j],a[p][j]);
      sp^=1;
    }
    for(int j=i+1;j<n;j++)
      if(a[j][i]){
    int t=1LL*a[j][i]*Pow(a[i][i],P-2)%P;
    for(int k=1;k<n;k++)
      a[j][k]=(a[j][k]+P-1LL*t*a[i][k]%P)%P;
      }
  }
  int ret=1; for(int i=1;i<n;i++) ret=1LL*ret*a[i][i]%P;
  if(sp) ret=-ret;
  return (ret+P)%P;
}

inline int calc(int x){
  static Mat tmp;
  memset(tmp,0,sizeof(tmp));
  for(int i=1;i<=n;i++)
    for(int j=i+1;j<=n;j++)
      if(i>cnt || j>cnt || (i<=x && j<=x)){
    tmp[i][j]=P-1; tmp[j][i]=P-1; tmp[i][i]++; tmp[j][j]++;
      }
  return det(tmp);
}

inline int C(int x,int y){
  return 1LL*fac[x]*inv[y]%P*inv[x-y]%P;
}

inline void part2(){
  for(int i=0;i<=cnt;i++) g[i]=calc(i);
  for(int i=1;i<=cnt;i++){
    for(int j=0;j<i;j++)
      g[i]=(g[i]-1LL*C(i,j)*g[j])%P;
    g[i]=(g[i]+P)%P;
  }
}

inline void Pre(){
  fac[0]=1; for(int i=1;i<=n;i++) fac[i]=1LL*fac[i-1]*i%P;
  inv[1]=1; for(int i=2;i<=n;i++) inv[i]=1LL*(P-P/i)*inv[P%i]%P;
  inv[0]=1; for(int i=1;i<=n;i++) inv[i]=1LL*inv[i]*inv[i-1]%P;
}

class SweetFruits{  
public:  
  int countTrees(vector <int> sweetness, int maxSweetness) {  
    n=sweetness.size(); lim=maxSweetness; Pre(); s.clear();
    for(int i=1;i<=n;i++){
      a[i]=sweetness[i-1];
      if(a[i]>=0) s.pb(a[i]);
    }
    cnt=s.size();
    memset(f,0,sizeof(f)); memset(g,0,sizeof(g));
    part1(); part2(); int ans=0;
    for(int i=0;i<=cnt;i++) ans=(ans+1LL*f[i]*g[i])%P;
    return ans;
  }


  // BEGIN CUT HERE
public:
  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(); if ((Case == -1) || (Case == 3)) test_case_3(); if ((Case == -1) || (Case == 4)) test_case_4(); }
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 Arr0[] = {1, 2, -1, 3}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 3; int Arg2 = 3; verify_case(0, Arg2, countTrees(Arg0, Arg1)); }
  void test_case_1() { int Arr0[] = {1, 2, -1, 3}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 5; int Arg2 = 7; verify_case(1, Arg2, countTrees(Arg0, Arg1)); }
  void test_case_2() { int Arr0[] = {-1, -1, 2, 5, 5}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 6; int Arg2 = 20; verify_case(2, Arg2, countTrees(Arg0, Arg1)); }
  void test_case_3() { int Arr0[] = {2, 6, 8, 4, 1, 10, -1, -1, -1, -1}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 15; int Arg2 = 17024000; verify_case(3, Arg2, countTrees(Arg0, Arg1)); }
  void test_case_4() { int Arr0[] = {1078451, -1, 21580110, 8284711, -1, 4202301, 3427559, 8261270, -1, 16176713,
                     22915672, 24495540, 19236, 5477666, 12280316, 3305896, 17917887, 564911, 22190488, 21843923,
                     23389728, 14641920, 9590140, 12909561, 20405638, 100184, 23336457, 12780498, 18859535, 23180993,
                     10278898, 5753075, 21250919, 17563422, 10934412, 22557980, 24895749, 7593671, 10834579, 5606562}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 245243285; int Arg2 = 47225123; verify_case(4, Arg2, countTrees(Arg0, Arg1)); }
  // END CUT HERE

};  

// BEGIN CUT HERE  
int main()  
{  
  SweetFruits ___test;
  ___test.run_test(4);
  system("pause");  
}  
// END CUT HERE