(Analysis by Danny Mittal, Benjamin Qi)

If the graph is bipartite, then the answer is simply the amount of ways for each node to have a nonzero amount of edges to nodes closer by $1$ edge to the source. Clearly, no other kinds of edges can be allowed. Letting $f_d$ be the amount of nodes at a distance $d$ from the source, this yields the answer

$$\prod_{d = 1}^{n - 1} \left(2^{f_{d - 1}} - 1\right)^{f_d}.$$

If the graph is not bipartite, then, as in Minimizing Edges from this contest and Counting Graphs from January, define

$$h(i)=(a_i,b_i)=(\min(dist_{even}(i),dist_{odd}(i)),\max(dist_{even}(i),dist_{odd}(i))).$$

For convenience, define $s(a, b)$ to be the set of all nodes $i$ with $h(i) = (a, b)$.

A graph having $f_G$ equivalent to that of the given graph must have edges adjacent to vertex $i$ satisfying at least one of the following conditions for each $2\le i\le N$ (the condition is slightly different for vertex $1$ since $a_1=0$).

1. $i$ is adjacent to a vertex in $s(a_i-1,b_i-1)$. Call this an edge of type A.
2. If $a_i+1<b_i$, then $i$ is adjacent to a vertex in $s(a_i-1,b_i+1)$ and a vertex in $s(a_i+1,b_i-1)$. Call these edges of type B.
3. If $a_i+1=b_i$, then $i$ is adjacent to a vertex in $s(a_i-1,b_i+1)$ and a vertex in $s(a_i,b_i)$ (possibly $i$ itself). Call an edge from $i$ to a vertex in $s(a_i, b_i)$ an edge of type C.

First observe that edges that are not type A, B, or C edges cannot exist in the graph. The problem is then to count the amount of ways to include edges of these types so that at least one of the above conditions is satisfied for each vertex.

As in Minimizing Edges, we can look at each layer (vertices with a fixed $a_i+b_i$) independently, as edges of type A are only relevant to the vertex in the higher layer, and edges of type B and C are between vertexes in the same layer.

We will then compute the answer for each layer using DP, similarly to the $\mathcal O(N^2)$ approach in Minimizing Edges.

We will define $dp_{a, b}(x)$ to be the amount of ways to choose edges in layer $a + b$ considering only the nodes $j$ in said layer with $a_j \leq a$ such that $x$ of the nodes in $s(a, b)$ do not yet have one of the above conditions satisfied. Our answer will be the product over all layers $l$ of $dp_{\frac {l - 1} 2, \frac {l + 1} 2}(0)$.

For each $(a, b)$, we will compute $dp_{a, b}$ in two steps. The first step will be to compute an intermediate DP $dp_{int}$, where we consider type B edges to nodes in $s(a - 1, b + 1)$ but don't yet consider type A edges to nodes in $s(a - 1, b - 1)$. $dp_{int}(x)$ will be defined similarly to above, except that here $x$ represents the amount of nodes in $s(a, b)$ with no edges at all yet, so that the other nodes in $s(a, b)$ aren't fully satisfied either but do at least have a type B edge to a node in $s(a - 1, b + 1)$.

To compute $dp_{int}(x)$ for all $x = 0, \ldots, |s_{a, b}|$, we transition from each $dp_{a - 1, b + 1}(y)$ for each $y = 0, \ldots, |s_{a - 1, b + 1}|$, considering how to add type B edges between nodes in $s(a - 1, b + 1)$ and nodes in $s(a, b)$. For a given $x, y$, we have $y$ nodes in $s(a - 1, b + 1)$ which we must add a type B edge to (we can, but don't have to, add type B edges to the other nodes in $s(a - 1, b + 1)$). Additionally, we have $|s(a, b)| - x$ nodes in $s(a, b)$ which we must add type B edges to and $x$ nodes in $s(a, b)$ which we cannot add type B edges to.

We can compute the amount of ways to transition using PIE (the principle of inclusion-exclusion). Consider the nodes in $s(a, b)$ which need an edge. We first add the amount of ways to add edges from each of these nodes to the nodes in $s(a - 1, b + 1)$ so that each of the nodes has at least one edge. This is $\left(2^{|s(a - 1, b + 1)|} - 1\right)^{|s(a, b)| - x}$. Noting that we must force the $y$ nodes in $s(a - 1, b + 1)$ to also have an edge, we then subtract the amount of ways that explicitly excludes one of these $y$ nodes. This is $y(2^{|s(a - 1, b + 1)| - 1} - 1)^{|s(a, b)| - x}$. Continuing with this PIE, we find that the amount of transitions is

$$\sum_{k = 0}^y (-1)^k \binom y k \left(2^{|s(a - 1, b + 1)|-k} - 1\right)^{|s(a, b)| - x}.$$

We perform this transition for each $y$. At the end, we also need to choose which $x$ nodes in $s(a, b)$ we're talking about, so we multiply our result by $\binom {|s(a, b)|} x$. This yields the overall formula

$$dp_{int}(x) = \binom {|s(a, b)|} x \sum_{y = 0}^{|s(a - 1, b + 1)|} dp_{a - 1, b + 1}(y) \sum_{k = 0}^y (-1)^k \binom y k \left(2^{|s(a - 1, b + 1)|-k} - 1\right)^{|s(a, b)| - x}.$$

The second step will be to finally compute $dp_{a, b}$ based on $dp_{int}$. Consider transitioning to $dp_{a, b}(x)$ from $dp_{int}(y)$. The $y$ nodes in $s(a, b)$, since they didn't get a type B edge to a node in $s(a - 1, b + 1)$, now urgently require a type A edge to a node in $s(a - 1, b - 1)$. Therefore, the $x$ nodes that don't get a type A edge and hence will not yet be satisfied must come from the other $|s(a, b)| - y$ nodes. We first choose these nodes, which we can do in $\binom {|s(a, b)| - y} x$ ways. Then, each of the other $|s(a, b) - x|$ nodes needs a nonzero amount of type A edges to nodes in $s(a - 1, b - 1)$. For each such node, we can choose these edges in $2^{|s(a - 1, b - 1)|} - 1$ ways. Hence, the amount of transitions is

$$\binom {|s(a, b)| - y} x \left(2^{|s(a - 1, b - 1)|} - 1\right)^{|s(a, b)| - x},$$

and so our overall formula is

$$dp_{a, b}(x) = \sum_{y = 0}^{|s(a, b)|} dp_{int}(y) \binom {|s(a, b)| - y} x \left(2^{|s(a - 1, b - 1)|} - 1\right)^{|s(a, b)| - x}.$$

There are two special cases: $a = 0$ and $a + 1 = b$. In the former case, we essentially don't do either step. In the latter case, we need a third step to calculate the actual value of $dp_{a, a + 1}(0)$, which must take into account type C edges. This can be done using PIE similarly to the first step, considering which of the $x$ nodes you explicitly exclude from having type C edges. The exact formula is left as an exercise to the reader.

The overall runtime of this algorithm is $\mathcal O(N^3)$ due to the first step, though solutions running in $\mathcal O(N^4)$ time with a good constant factor were sufficient for full credit.

Here is Ben's code (which uses the exact same formulas as described above):

#include <bits/stdc++.h>

using namespace std;

const int MOD = 1e9+7;

using ll = long long;

#define f first
#define s second

struct mi {
int v;
explicit operator int() const { return v; }
mi() { v = 0; }
mi(ll _v):v(int(_v%MOD)) { v += (v<0)*MOD; }
};
mi& operator+=(mi& a, mi b) {
if ((a.v += b.v) >= MOD) a.v -= MOD;
return a; }
mi& operator-=(mi& a, mi b) {
if ((a.v -= b.v) < 0) a.v += MOD;
return a; }
mi operator+(mi a, mi b) { return a += b; }
mi operator-(mi a, mi b) { return a -= b; }
mi operator*(mi a, mi b) { return mi((ll)a.v*b.v); }
mi& operator*=(mi& a, mi b) { return a = a*b; }
mi pow(mi a, ll p) { assert(p >= 0); // asserts are important!
return p==0?1:pow(a*a,p/2)*(p&1?a:1); }

using vi = vector<int>;
using vmi = vector<mi>;

int N,M;

vector<array<int,2>> gen_dist() { // BFS
vector<array<int,2>> dist(N,{MOD,MOD});
queue<pair<int,int>> q;
auto ad = [&](int a, int b) {
if (dist[a][b%2] != MOD) return;
dist[a][b%2] = b; q.push({a,b});
};
while (!q.empty()) {
pair<int,int> p = q.front(); q.pop();
}
return dist;
}

mi comb[105][105]; // comb[x][y] = (x choose y)
vmi p2; // stores powers of 2

void solve() {
cin >> N >> M;
for (int i = 0; i < M; ++i) {
int a,b; cin >> a >> b; --a,--b;
}
vector<array<int,2>> dists = gen_dist();
for (array<int,2>& t: dists)
if (t[0] > t[1]) swap(t[0],t[1]);
if (dists[0][1] == MOD) { // bipartite
vi num_at_dist(N);
for (int i = 0; i < N; ++i)
++num_at_dist[min(dists[i][0],dists[i][1])];
mi ans = 1;
for (int i = 1; i < N; ++i)
ans *= pow(p2[num_at_dist[i-1]]-1,num_at_dist[i]);
cout << (int)ans << "\n";
return;
}
vector<vi> S(4*N,vi(4*N));
for (array<int,2> t: dists) ++S[t[0]][t[1]];
mi ans = 1;
for (int sum = 1; sum < 4*N; sum += 2) {
vmi dp(S[0][sum]+1); dp.back() = 1;
for (int a = 1; a <= sum/2; ++a) { // solve in increasing order of a
int b = sum-a, num = S[a][b];
vmi dp_int(num+1);
{ // deal with s_{a-1,b+1} to s_{a,b}
int prev_num = S[a-1][b+1];
for (int x = 0; x <= num; ++x) { // x in s_{a,b} with no edges at all, num-x that receive edges
for (int y = 0; y <= prev_num; ++y) { // y nodes in s_{a-1,b+1} that need an edge
mi tot_mul = 0;
for (int k = 0; k <= y; ++k) { // suppose that k out of y didn't actually get an edge
mi mul = comb[y][k]*pow(p2[prev_num-k]-1,num-x);
if (k&1) tot_mul -= mul;
else tot_mul += mul;
}
dp_int[x] += tot_mul*dp[y];
}
dp_int[x] *= comb[num][x];
}
}
{ // deal with s_{a-1,b-1} to s_{a,b}
dp = vmi(num+1);
int lower_num = S[a-1][b-1];
for (int x = 0; x <= num; ++x) { // x will not be part of such an edge
for (int y = 0; x+y <= num; ++y) { // y urgently need an edge, num-y don't
mi mul = comb[num-y][x]*pow(p2[lower_num]-1,num-x);
dp[x] += mul*dp_int[y];
}
}
}
}
// finally, deal with clique edges
mi ans_for_layer = 0;
int num = S[sum/2][sum/2+1];
for (int y = 0; y <= num; ++y) { // how many ways to complete if y need an edge
mi tot_mul = 0;
for (int k = 0; k <= y; ++k) {
mi mul = comb[y][k]*p2[(num-k)*(num-k+1)/2];
if (k&1) tot_mul -= mul;
else tot_mul += mul;
}
ans_for_layer += tot_mul*dp[y];
}
ans *= ans_for_layer;
}
cout << (int)ans << "\n";
}

int main() {
comb[0][0] = 1;
for (int i = 1; i < 105; ++i) {
comb[i][0] = 1;
for (int j = 1; j <= i; ++j)
comb[i][j] = comb[i-1][j]+comb[i-1][j-1];
}
p2 = {1};
for (int _ = 0; _ < 10000; ++_)
p2.push_back(2*p2.back());
int TC; cin >> TC;
while (TC--) solve();
}


Danny's code:

import java.io.BufferedReader;
import java.io.IOException;
import java.util.*;

public class CountingGraphs4FixedBounds {
public static final long MOD = 1000000007L;

public static void main(String[] args) throws IOException {
for (int tc = 1; tc <= t; tc++) {
int n = Integer.parseInt(tokenizer.nextToken());
int m = Integer.parseInt(tokenizer.nextToken());
List<Integer>[] adj = new List[(2 * n) + 1];
for (int a = 1; a <= 2 * n; a++) {
}
for (int j = 1; j <= m; j++) {
int a = Integer.parseInt(tokenizer.nextToken());
int b = Integer.parseInt(tokenizer.nextToken());
}
int[] dist = new int[(2 * n) + 1];
Arrays.fill(dist, -1);
dist[1] = 0;
while (!q.isEmpty()) {
int a = q.remove();
for (int b : adj[a]) {
if (dist[b] == -1) {
dist[b] = dist[a] + 1;
}
}
}
long[] pow2 = new long[(n * n) + 1];
pow2[0] = 1;
for (int j = 1; j <= n * n; j++) {
pow2[j] = (2L * pow2[j - 1]) % MOD;
}
long[][] powMersenne = new long[n + 1][n + 1];
for (int a = 0; a <= n; a++) {
powMersenne[a][0] = 1L;
for (int j = 1; j <= n; j++) {
powMersenne[a][j] = ((pow2[a] - 1L) * powMersenne[a][j - 1]) % MOD;
}
}
long[][] choose = new long[n + 1][n + 1];
for (int a = 0; a <= n; a++) {
choose[a][0] = 1;
for (int b = 1; b <= a; b++) {
choose[a][b] = (choose[a - 1][b - 1] + choose[a - 1][b]) % MOD;
}
}
if (dist[n + 1] == -1) {
int[] freq = new int[n];
for (int a = 1; a <= n; a++) {
freq[Math.max(dist[a], dist[n + a])]++;
}
for (int d = 1; d < n; d++) {
}
} else {
int[][] freq = new int[2 * n][2 * n];
for (int a = 1; a <= n; a++) {
freq[Math.min(dist[a], dist[n + a])][Math.max(dist[a], dist[n + a])]++;
}
for (int s = 1; s <= (4 * n) - 2; s += 2) {
long[] dp = new long[0];
for (int j = s / 2, k = j + 1; j >= 0 && k < 2 * n; j--, k++) {
long[] dp1 = new long[freq[j][k] + 1];
if (j == s / 2) {
for (int a = 0; a <= freq[j][k]; a++) {
for (int b = 0; b <= a; b++) {
dp1[a] += (a % 2 == b % 2 ? 1L : -1L) * choose[a][b] * pow2[(b * (b + 1)) / 2];
dp1[a] %= MOD;
}
dp1[a] *= choose[freq[j][k]][a];
dp1[a] %= MOD;
}
} else {
for (int a = 0; a <= freq[j + 1][k - 1]; a++) {
for (int b = 0; b <= freq[j][k]; b++) {
long ways = 0;
for (int c = 0; c <= freq[j + 1][k - 1] - a; c++) {
ways += ((freq[j + 1][k - 1] - a) % 2 == c % 2 ? 1L : -1L) * choose[freq[j + 1][k - 1] - a][c] * powMersenne[a + c][b];
ways %= MOD;
}
dp1[b] += ((choose[freq[j][k]][b] * dp[a]) % MOD) * ways;
dp1[b] %= MOD;
}
}
}
if (j == 0) {
dp = dp1;
} else {
long[] dp2 = new long[freq[j][k] + 1];
for (int a = 0; a <= freq[j][k]; a++) {
for (int b = freq[j][k] - a; b <= freq[j][k]; b++) {
dp2[b] += ((choose[a][b - (freq[j][k] - a)] * powMersenne[freq[j - 1][k - 1]][b]) % MOD) * dp1[a];
dp2[b] %= MOD;
}
}
dp = dp2;
}
}
}
}
}
}
}


In fact, the PIE can be simplified to give a $\mathcal{O}(N^2)$ solution as demonstrated by Andrew He's code here. It helps to think of the number of edge covers of a set of vertices $S$ as

$$\sum_{T\subseteq S}(-1)^T2^{(\#\text{ of edges within }T)}.$$

Here is the code from the first solution above, slightly modified to run in $\mathcal{O}(N^2)$:

#include <bits/stdc++.h>

using namespace std;

const int MOD = 1e9+7;

using ll = long long;

#define f first
#define s second

struct mi {
int v;
explicit operator int() const { return v; }
mi() { v = 0; }
mi(ll _v):v(int(_v%MOD)) { v += (v<0)*MOD; }
};
mi& operator+=(mi& a, mi b) {
if ((a.v += b.v) >= MOD) a.v -= MOD;
return a; }
mi& operator-=(mi& a, mi b) {
if ((a.v -= b.v) < 0) a.v += MOD;
return a; }
mi operator+(mi a, mi b) { return a += b; }
mi operator-(mi a, mi b) { return a -= b; }
mi operator*(mi a, mi b) { return mi((ll)a.v*b.v); }
mi& operator*=(mi& a, mi b) { return a = a*b; }
mi pow(mi a, ll p) { assert(p >= 0); // asserts are important!
return p==0?1:pow(a*a,p/2)*(p&1?a:1); }

using vi = vector<int>;
using vmi = vector<mi>;

int N,M;

vector<array<int,2>> gen_dist() { // BFS
vector<array<int,2>> dist(N,{MOD,MOD});
queue<pair<int,int>> q;
auto ad = [&](int a, int b) {
if (dist[a][b%2] != MOD) return;
dist[a][b%2] = b; q.push({a,b});
};
while (!q.empty()) {
pair<int,int> p = q.front(); q.pop();
}
return dist;
}

mi comb[105][105]; // comb[x][y] = (x choose y)
vmi p2; // stores powers of 2

void solve() {
cin >> N >> M;
for (int i = 0; i < M; ++i) {
int a,b; cin >> a >> b; --a,--b;
}
vector<array<int,2>> dists = gen_dist();
for (array<int,2>& t: dists)
if (t[0] > t[1]) swap(t[0],t[1]);
if (dists[0][1] == MOD) { // bipartite
vi num_at_dist(N);
for (int i = 0; i < N; ++i)
++num_at_dist[min(dists[i][0],dists[i][1])];
mi ans = 1;
for (int i = 1; i < N; ++i)
ans *= pow(p2[num_at_dist[i-1]]-1,num_at_dist[i]);
cout << (int)ans << "\n";
return;
}
vector<vi> S(4*N,vi(4*N));
for (array<int,2> t: dists) ++S[t[0]][t[1]];
mi ans = 1;
for (int sum = 1; sum < 4*N; sum += 2) {
vmi dp(S[0][sum]+1); dp.back() = 1;
for (int a = 1; a <= sum/2; ++a) { // solve in increasing order of a
int b = sum-a, num = S[a][b];
vmi dp_int(num+1);
{ // deal with s_{a-1,b+1} to s_{a,b}
int prev_num = S[a-1][b+1];
for (int y = prev_num; y >= 0; --y)
for (int y2 = y+1; y2 <= prev_num; ++y2) {
// only y nodes from s_{a-1,b+1} need edges to s_{a,b}
// say y2 >= y nodes from s_{a-1,b+1} actually have edges to s_{a,b}
dp[y2] += comb[prev_num-y][y2-y]*dp[y];
}
// now, dp[y] -> y nodes from s_{a-1,b+1} that must have edges to s_{a,b},
// rest of nodes from s_{a-1,b-1} don't

for (int y = 0; y <= prev_num; ++y)
for (int x = 0; x <= num; ++x)
dp_int[x] += dp[y]*pow(p2[x]-1,y);
// now, dp_int[x] stores the number of ways to satisfy all nodes from s_{a-1,b+1}
// by drawing edges to x nodes in s_{a,b}

// can replace the three lines above with the commented out lines below instead
// which in turn can be done in O(N log N) with Chirp-Z Transform
// (https://cp-algorithms.com/algebra/polynomial.html)
// for (int y = 0; y <= prev_num; ++y)
// 	for (int y2 = 0; y2 < y; ++y2) {
// 		if ((y-y2)&1) dp[y2] -= bad;
// 	}
// for (int y = 0; y <= prev_num; ++y)
// 	for (int x = 0; x <= num; ++x)
// 		dp_int[x] += dp[y]*pow(p2[x],y);

for (int x = num; x >= 0; --x) // apply PIE again
for (int xx = x-1; xx >= 0; --xx) {
}
reverse(begin(dp_int),end(dp_int));
for (int x = 0; x <= num; ++x) dp_int[x] *= comb[num][x];
}
{ // deal with s_{a-1,b-1} to s_{a,b}
dp = vmi(num+1);
int lower_num = S[a-1][b-1];
for (int x = 0; x <= num; ++x) { // x will not be part of such an edge
for (int y = 0; x+y <= num; ++y) { // y urgently need an edge, num-y don't
mi mul = comb[num-y][x]*pow(p2[lower_num]-1,num-x);
dp[x] += mul*dp_int[y];
}
}
}
}
// finally, deal with clique edges
mi ans_for_layer = 0;
int num = S[sum/2][sum/2+1];
for (int y = 0; y <= num; ++y) { // how many ways to complete if y need an edge
mi tot_mul = 0;
for (int k = 0; k <= y; ++k) {
mi mul = comb[y][k]*p2[(num-k)*(num-k+1)/2];
if (k&1) tot_mul -= mul;
else tot_mul += mul;
}
ans_for_layer += tot_mul*dp[y];
}
ans *= ans_for_layer;
}
cout << (int)ans << "\n";
}

int main() {
comb[0][0] = 1;
for (int i = 1; i < 105; ++i) {
comb[i][0] = 1;
for (int j = 1; j <= i; ++j)
comb[i][j] = comb[i-1][j]+comb[i-1][j-1];
}
p2 = {1};
for (int _ = 0; _ < 10000; ++_)
p2.push_back(2*p2.back());
int TC; cin >> TC;
while (TC--) solve();
}


In fact, this can be sped up to $\mathcal O(N\log N)$ using FFT to quickly multiply polynomials.