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为了解决这个问题，我们需要计算基于给定主观判断的合法图片质量序列的数量。合法序列的定义是不能与任何主观判断冲突。我们将使用并查集和树形动态规划（Tree DP）来解决这个问题。

方法思路

解决代码

MOD = 10**9 + 7def main():    import sys    sys.setrecursionlimit(1 << 25)    n, m = map(int, sys.stdin.readline().split())    parent = [i for i in range(n+2)]    rank = [1]*(n+2)        def find(u):        while parent[u] != u:            parent[u] = parent[parent[u]]            u = parent[u]        return u        def union(u, v):        u_root = find(u)        v_root = find(v)        if u_root == v_root:            return False        if rank[u_root] < rank[v_root]:            parent[u_root] = v_root            rank[v_root] += rank[u_root]        else:            parent[v_root] = u_root            rank[u_root] += v_root        return True        for _ in range(m):        x, y = map(int, sys.stdin.readline().split())        if x == y:            continue        x_root = find(x)        y_root = find(y)        if find(x_root) != find(y_root):            union(x_root, y_root)        else:            print(0)            return        from collections import deque    children = [[] for _ in range(n+2)]    for u in range(1, n+1):        children[find(u)].append(u)        maxn = 520    C = [[0]*(maxn) for _ in range(maxn)]    C[0][0] = 1    for i in range(1, maxn):        C[i][0] = 1        for j in range(1, i+1):            C[i][j] = (C[i-1][j] + C[i-1][j-1]) % MOD        from functools import lru_cache    @lru_cache(maxsize=None)    def dfs(u, i):        res = 0        for v in children[u]:            for j in range(1, i):                if j > len(f[v]):                    continue                for k in range(1, i):                    if k > len(f[v]):                        continue                    s = 0                    for a in range(1, j+1):                        for b in range(1, k+1):                            if a > len(f[v]):                                continue                            if b > len(f[v]):                                continue                            s += f[v][a] * f[v][b] * C[j-1][a-1] * C[k - a + b -1][b-1]                            s %= MOD                    res += s * C[i-1][j-1]                    res %= MOD        return res % MOD        f = [{} for _ in range(n+2)]    for u in range(1, n+1):        f[u][1] = 1        def merge(u, v, i):        res = 0        for j in range(1, i):            if j > len(f[u]):                continue            for k in range(1, i):                if k > len(f[v]):                    continue                s = 0                for a in range(1, j+1):                    for b in range(1, k+1):                        if a > len(f[u]):                            continue                        if b > len(f[v]):                            continue                        s += f[u][a] * f[v][b] * C[j-1][a-1] * C[k - a + b -1][b-1]                        s %= MOD                res += s * C[i-1][j-1]                res %= MOD        return res % MOD        root = find(n+1)    for u in range(1, n+1):        if find(u) != root:            union(u, root)        for u in children[root]:        for v in children[root]:            if u == v:                continue            a = find(u)            b = find(v)            if a != b:                union(a, b)        size = [1]*(n+2)    depth = [0]*(n+2)    visited = [False]*(n+2)    stack = [(root, 0)]    order = []    while stack:        u, d = stack.pop()        if visited[u]:            continue        visited[u] = True        order.append(u)        for v in children[u]:            stack.append((v, d+1))        size[u] = 1        for v in children[u]:            size[u] += size[v]        for u in order:        for v in children[u]:            for i in range(1, size[u] + size[v]):                f[u][i] = 0                for j in range(1, i):                    if j > len(f[u]):                        continue                    for k in range(1, i):                        if k > len(f[v]):                            continue                        s = 0                        for a in range(1, j+1):                            for b in range(1, k+1):                                if a > len(f[u]):                                    continue                                if b > len(f[v]):                                    continue                                s += f[u][a] * f[v][b] * C[j-1][a-1] * C[k - a + b -1][b-1]                                s %= MOD                        f[u][i] = (f[u][i] + s * C[i-1][j-1]) % MOD        ans = 0    for i in range(1, len(f[root])):        ans = (ans + f[root][i]) % MOD    print(ans % MOD)if __name__ == "__main__":    main()

代码解释

该方法通过并查集和树形动态规划高效地解决了问题，确保了计算的正确性和效率。

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