[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

;
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

;
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
;
int S;

ll tree
;
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]++;
else
a[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
;
vector<abcd> lft;
vector<ll> rgt
;
int pt
;

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