題目鏈接
hdu5279
題解
給出若干個完全圖，然后完全圖之間首尾相連并成環(huán)，要求刪邊使得兩點之間路徑數(shù)不超過\(1\)，求方案數(shù)
容易想到各個完全圖是獨立的，每個完全圖要刪成一個森林，其實就是詢問\(n\)個點有標號森林的個數(shù)
設(shè)\(f[i]\)表示\(i\)個點有標號森林的個數(shù)
枚舉第一個點所在樹大小，我們只需求出\(n\)個點有多少種樹，由\(purfer\)序容易知道是\(n^{n - 2}\)
那么有
\[f[n] = \sum\limits_{i = 1}^{n} {n - 1 \choose i - 1}i^{i - 2}f[n - i]\]
化簡一下：
\[f[n] = (n - 1)!\sum\limits_{i = 1}^{n}\frac{i^{i - 2}}{(i - 1)!} \times \frac{f[n - i]}{(n - i)!}\]
分治\(NTT\)即可
每個完全圖的方案是\(f[a[i]]\)，中間相連的\(n\)條邊有\(2^n\)種方案，由乘法原理乘起來即可
但是這樣求出來的不是答案，會多算一類情況：
每個完全圖的\(1\)和\(a_i\)相通且所有中介邊存在
所以我們還需要計算\(g[i]\)表示\(i\)個點的森林，\(1\)和\(i\)點在同一棵樹內(nèi)的方案數(shù)
顯然
\[g[n] = \sum\limits_{i = 2}^{n} {n - 2 \choose i - 2}i^{i - 2}f[n - i]\]
化簡得
\[g[n] = (n - 2)!\sum\limits_{i = 2}^{n} \frac{i^{i - 2}}{(i - 2)!} \times \frac{f[n - i]}{(n - i)!}\]
\(NTT\)即可
最后答案減去\(g[a[i]]\)的乘積即可
復雜度\(O(nlog^2n)\)
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 400005,maxm = 100005,INF = 1000000000;
inline int read(){int out = 0,flag = 1; char c = getchar();while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}return out * flag;
}
const int G = 3,P = 998244353;
int R[maxn];
inline int qpow(int a,int b){int re = 1;for (; b; b >>= 1,a = 1ll * a * a % P)if (b & 1) re = 1ll * re * a % P;return re;
}
void NTT(int* a,int n,int f){for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);for (int i = 1; i < n; i <<= 1){int gn = qpow(G,(P - 1) / (i << 1));for (int j = 0; j < n; j += (i << 1)){int g = 1,x,y;for (int k = 0; k < i; k++,g = 1ll * g * gn % P){x = a[j + k],y = 1ll * g * a[j + k + i] % P;a[j + k] = (x + y) % P,a[j + k + i] = ((x - y) % P + P) % P;}}}if (f == 1) return;int nv = qpow(n,P - 2); reverse(a + 1,a + n);for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % P;
}int f[maxn],g[maxn],fac[maxn],fv[maxn],p[maxn],N = 100005;
int A[maxn],B[maxn];
void solve(int l,int r){if (l == r){if (l > 0) f[l] = 1ll * f[l] * fac[l - 1] % P;return;}int mid = l + r >> 1;solve(l,mid);int n,m,L;m = mid - l + 1;for (int i = 0; i < m; i++) A[i] = 1ll * f[l + i] * fv[l + i] % P;m = r - l;for (int i = 0; i < m; i++) B[i] = 1ll * p[i + 1] * fv[i] % P;n = 1; L = 0; m = mid + r - (l << 1) - 1;while (n <= m) n <<= 1,L++;for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));for (int i = mid - l + 1; i < n; i++) A[i] = 0;for (int i = r - l; i < n; i++) B[i] = 0;NTT(A,n,1); NTT(B,n,1);for (int i = 0; i < n; i++) A[i] = 1ll * A[i] * B[i] % P;NTT(A,n,-1);for (int i = mid - l,j = mid + 1; j <= r; i++,j++){f[j] = (f[j] + A[i]) % P;}solve(mid + 1,r);
}
int b[maxn];
inline int C(int n,int m){if (m > n) return 0;return 1ll * fac[n] * fv[m] % P * fv[n - m] % P;
}
void work(){fac[0] = p[0] = p[1] = 1;for (int i = 1; i <= N + 2; i++)fac[i] = 1ll * fac[i - 1] * i % P;for (int i = 2; i <= N + 2; i++)p[i] = qpow(i,i - 2);fv[N + 2] = qpow(fac[N + 2],P - 2); fv[0] = 1;for (int i = N + 1; i; i--)fv[i] = 1ll * fv[i + 1] * (i + 1) % P;f[0] = 1;solve(0,N);A[0] = A[1] = 0;for (int i = 2; i <= N; i++) A[i] = 1ll * p[i] * fv[i - 2] % P;for (int i = 0; i <= N; i++) B[i] = 1ll * f[i] * fv[i] % P;int n = 1,L = 0;while (n <= (N << 1)) n <<= 1,L++;for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));for (int i = N + 1; i < n; i++) A[i] = 0;for (int i = N + 1; i < n; i++) B[i] = 0;NTT(A,n,1); NTT(B,n,1);for (int i = 0; i < n; i++) A[i] = 1ll * A[i] * B[i] % P;NTT(A,n,-1);for (int i = 2; i <= N; i++) g[i] = 1ll * A[i] * fac[i - 2] % P;g[1] = 1;
}
int n,a[maxn],ans,ans2;
int main(){work();//REP(i,100) printf("%d ",f[i]); puts("");//REP(i,100) printf("%d ",g[i]); puts("");int T = read();while (T--){n = read();REP(i,n) a[i] = read();ans = qpow(2,n);REP(i,n) ans = 1ll * ans * f[a[i]] % P;ans2 = 1;REP(i,n) ans2 = 1ll * ans2 * g[a[i]] % P;ans = ((ans - ans2) % P + P) % P;printf("%d\n",ans);}return 0;
}
轉(zhuǎn)載于:https://www.cnblogs.com/Mychael/p/9172482.html
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