LOJ#6072. 「2017 山东一轮集训 Day5」苹果树解题报告

好苹果会组成连通块，整棵树的权值为

∑ i = 1 n c i [ c i ≥ 0 ] [ s i z n u m ( c i ) > 1 ] \sum_{i=1}^nc_i[c_i\ge 0][siz_{num(c_i)}>1] i=1∑nci[ci≥0][siznum(ci)>1]

设原树中有 m m m 个好苹果，当有 k k k 个好苹果计入权值（即形成了 s i z > 1 siz>1 siz>1 的连通块）， m − k m-k m−k 个好苹果不计入权值（即是孤点），我们求生成树个数。看到生成树我们会想到 Matrix-Tree 定理，不妨试着构造一下。

一个新的 n n n 个点的图，其中 1 ∼ k 1\sim k 1∼k 表示形成连通块的好苹果（PART 1）， k + 1 ∼ m k+1\sim m k+1∼m 表示孤点的好苹果（PART 2）， m + 1 ∼ n m+1\sim n m+1∼n 表示坏苹果（PART 3）。

连以下几种边：

PART 1 ~ PART 1
PART 1 ~ PART 3
PART 2 ~ PART 3
PART 3 ~ PART 3

我们不能保证 PART 1 的连通块大小都 > 1 > 1 >1，也就是说，求出的答案其实是 ≤ k \le k ≤k 个好苹果形成连通块的 s i z > 1 siz>1 siz>1 的方案，记为 g ( k ) g(k) g(k)。恰好 k k k 个好苹果形成连通块的 s i z > 1 siz>1 siz>1 的方案，记为 f ( k ) f(k) f(k)，则

g ( k ) = ∑ i = 0 k ( k i ) f ( i ) g(k)=\sum_{i=0}^k\binom{k}{i}f(i) g(k)=i=0∑k(ik)f(i)

二项式反演，得到

f ( k ) = ∑ i = 0 k ( − 1 ) k − i ( k i ) g ( i ) f(k)=\sum_{i=0}^k(-1)^{k-i}\binom{k}{i}g(i) f(k)=i=0∑k(−1)k−i(ik)g(i)

于是题目的关键转为如何求计入权值的好苹果数量为 k k k、权值 ≤ l i m \le lim ≤lim 的选点方案数。这个是个经典问题，可以折半搜索+双指针解决。

#include<cstring>
#include<algorithm>
#include<cstdio>
#include<vector>
using namespace std;
typedef long long ll;
char In[1000000], *ptrs = In, *ptrt = In;
#define getchar() (ptrs == ptrt && (ptrt = (ptrs = In) + fread(In, 1, 1000000, stdin), ptrs == ptrt) ? EOF : *ptrs++)
ll read() {ll x = 0, f = 1; char ch = getchar();for(; ch < '0' || ch > '9'; ch = getchar()) if(ch == '-') f = -1;for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + int(ch - '0');return x * f;
}
const int MAXN = 45, P = 1e9 + 7;
namespace MINT {struct mint {int v;mint(int v = 0) : v(v) {}};int MOD(int v) {return v >= P ? v - P : v;}mint operator + (mint a, mint b) {return MOD(a.v + b.v);}mint operator - (mint a, mint b) {return MOD(a.v - b.v + P);}mint operator * (mint a, mint b) {return 1ll * a.v * b.v % P;}mint qpow(mint a, int n=P-2) {mint ret = 1; for(; n; n >>= 1, a = a * a) if(n & 1) ret = ret * a; return ret;}mint operator += (mint& a, mint b) {return a = a + b;}mint operator -= (mint& a, mint b) {return a = a - b;}mint operator *= (mint& a, mint b) {return a = a * b;}mint operator ++ (mint& a) {return a = a + 1;}mint operator -- (mint& a) {return a = a - 1;}
} using namespace MINT;
int n, m, lim, c[MAXN], a[MAXN];
vector<int> pres[MAXN], sufs[MAXN];
mint h[MAXN];
void calch() {for(int s = 0; s < (1 << (m/2)); s++) {int cnt = 0, sum = 0;for(int i = 0; i < m/2; i++)if((s >> i) & 1) {cnt++; sum += a[i+1];}pres[cnt].push_back(sum);}for(int s = 0; s < (1 << (m-m/2)); s++) {int cnt = 0, sum = 0;for(int i = 0; i < m-m/2; i++)if((s >> i) & 1) {cnt++; sum += a[i+1+m/2];}sufs[cnt].push_back(sum);}for(int i = 0; i <= m/2; i++) {if(pres[i].empty()) continue;sort(pres[i].begin(), pres[i].end());for(int j = 0; j <= m-m/2; j++) {if(sufs[j].empty()) continue;sort(sufs[j].begin(), sufs[j].end());vector<int> &A = pres[i], &B = sufs[j];int q = B.size()-1; mint ans = 0;for(int p = 0; p < (int)A.size() && q >= 0; p++) {while(q >= 0 && A[p] + B[q] > lim) q--;if(q < 0) break;ans += q+1;}h[i+j] += ans;}}
}
mint C[MAXN][MAXN], f[MAXN], g[MAXN];
mint e[MAXN][MAXN];
void addedge(int u, int v) {--u, --v; ++e[u][u]; ++e[v][v]; --e[u][v]; --e[v][u];}//delete the line 1.
mint det(mint a[][MAXN], int n) {int f = 0;for(int i = 1; i <= n; i++) {int num = -1;for(int j = i; j <= n; j++)if(a[j][i].v) { num = j; break; }if(num == -1) return 0;if(i != num) {f ^= 1;for(int j = i; j <= n; j++) swap(a[i][j], a[num][j]);}mint in = qpow(a[i][i]);for(int j = i+1; j <= n; j++) {mint t = in * a[j][i];for(int k = i; k <= n; k++) a[j][k] -= a[i][k] * t;}}mint ans = 1; for(int i = 1; i <= n; i++) ans *= a[i][i];if(f) ans = 0-ans;return ans;
}
mint calcg(int k) {for(int i = 1; i <= n-1; i++)for(int j = 1; j <= n-1; j++)e[i][j] = 0;for(int i = 1; i <= k; i++) {for(int j = i+1; j <= k; j++) addedge(i, j);for(int j = m+1; j <= n; j++) addedge(i, j);}for(int i = k+1; i <= m; i++)for(int j = m+1; j <= n; j++)addedge(i, j);for(int i = m+1; i <= n; i++)for(int j = i+1; j <= n; j++) addedge(i, j);return det(e, n-1);
}
int main() {n = read(), lim = read();for(int i = 1; i <= n; i++) {c[i] = read();if(c[i] != -1) a[++m] = c[i];}calch();for(int i = 0; i <= m; i++) g[i] = calcg(i);C[0][0] = 1;for(int i = 1; i <= m; i++) {C[i][0] = 1;for(int j = 1; j <= i; j++) C[i][j] = C[i-1][j-1] + C[i-1][j];}for(int i = 0; i <= m; i++) {f[i] = 0;for(int j = 0; j <= i; j++)if((i - j) & 1) f[i] -= C[i][j] * g[j];else f[i] += C[i][j] * g[j];}mint ans = 0;for(int i = 0; i <= m; i++) ans += h[i] * f[i];printf("%d\n", ans.v);return 0;
}