[meet in middle 矩阵树定理容斥原理] SRM 551 div1 SweetFruits

集训队论文传送门

大概就是我们先用meet in middle求出有恰好k个真甜的方案数
然后我们求这些东西的生成树个数乘在一起的和就是答案
我们让真甜连真甜真甜连不甜假甜连不甜不甜连不甜跑一发矩阵树定理
这样只能保证这些真甜的某个子集是真甜那么我们需要用 0~k-1 的简单容斥一下

// 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>
#include<functional>
#define pb push_back
#define cl(x) memset(x,0,sizeof(x))
using namespace std;
typedef long long ll;
typedef pair<ll,int> abcd;const int N=45;
const int P=1e9+7;ll C[N][N];
inline void Pre(int n){C[0][0]=1;for (int i=1;i<=n;i++){C[i][0]=1;for (int j=1;j<=i;j++)C[i][j]=(C[i-1][j]+C[i-1][j-1])%P;}
}inline ll Pow(ll a,int b){ll ret=1;for (;b;b>>=1,a=a*a%P)if (b&1)ret=ret*a%P;return ret;
}
inline ll Inv(ll a){return Pow(a,P-2);
}int a[N][N];
inline int det(int n){int f=0;for (int i=1;i<=n;i++){int k=0;for (int j=i;j<=n;j++) if (a[j][i]) {k=j; break; }if (i^k) { for (int j=i;j<=n;j++) swap(a[i][j],a[k][j]); f^=1; }for (int j=i+1;j<=n;j++){ll t=(ll)Inv(a[i][i])*a[j][i]%P;for (int k=i;k<=n;k++)(a[j][k]+=P-(ll)t*a[i][k]%P)%=P;}}ll ret=1;for (int i=1;i<=n;i++) ret=ret*a[i][i]%P;return f?(P-ret)%P:ret;
}int n,Maxs,val[N];
int S;ll tree[N];
inline void Calc(int k){for (int i=1;i<=n;i++) a[i][i]=0;for (int i=1;i<=n;i++)for (int j=i+1;j<=n;j++)if (!(j>k && j<=S))a[i][j]=a[j][i]=P-1,a[i][i]++,a[j][j]++;elsea[i][j]=a[j][i]=0;tree[k]=det(n-1);for (int i=0;i<k;i++)tree[k]+=P-(ll)C[k][i]*tree[i]%P;tree[k]%=P;
}ll cnt[N];
vector<abcd> lft;
vector<ll> rgt[N];
int pt[N];inline void Count(int n){lft.clear();for (int i=0;i<=n-n/2;i++) rgt[i].clear();for (int i=0;i<(1<<(n/2));i++){int c=0; ll s=0;for (int j=0;j<n/2;j++) if (i>>j&1) c++,s+=val[j+1];lft.pb(abcd(s,c));}sort(lft.begin(),lft.end());for (int i=0;i<(1<<(n-n/2));i++){int c=0; ll s=0;for (int j=0;j<n-n/2;j++) if (i>>j&1) c++,s+=val[n/2+j+1];rgt[c].pb(s);}for (int i=0;i<=n-n/2;i++)sort(rgt[i].begin(),rgt[i].end()),pt[i]=rgt[i].size()-1;for (int i=0;i<=n;i++) cnt[i]=0; for (int i=0;i<(int)lft.size();i++){ll s=lft[i].first; int c=lft[i].second;for (int j=0;j<=n-n/2;j++){while (~pt[j] && s+rgt[j][pt[j]]>Maxs)pt[j]--;cnt[j+c]+=pt[j]+1;}}for (int i=0;i<=n;i++) cnt[i]%=P;
}inline int Solve(){ll ans=0;S=0; Pre(n);for (int i=1;i<=n;i++) if (~val[i]) S++;sort(val+1,val+n+1,greater<int>());for (int k=0;k<=S;k++) Calc(k);Count(S);for (int k=0;k<=S;k++)ans+=cnt[k]*tree[k]%P;return ans%P;
}class SweetFruits{
public:int countTrees(vector <int> sweetness, int maxSweetness){Maxs=maxSweetness; n=sweetness.size();for (int i=1;i<=n;i++) val[i]=sweetness[i-1];return Solve();}// 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(-1);getch() ;return 0;
}
// END CUT HERE

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