As @Emre suggested, if you already have the distribution of topics in each document, you can represent each document as a vector $x_d \in \mathbb{R}^N$, where $N$ is the number of the unique topics in your collection. For documents, not exhibiting specific topics just fill the specific cells in each feature vector with zeros. Then, you can use some clustering algorithm such as nearest neigbors, using those feature vectors.
Example usage code in python below:
import pandas as pd
import numpy as np
from sklearn.metrics import pairwise_distances
# Initialize some documents
doc1 = {'Science':0.7, 'History':0.05, 'Politics':0.15, 'Sports':0.1}
doc2 = {'News':0.3, 'Art':0.5, 'Politics':0.1, 'Sports':0.1}
doc3 = {'Science':0.8, 'History':0.1, 'Politics':0.05, 'News':0.1}
doc4 = {'Science':0.2, 'Weather':0.2, 'Art':0.6, 'Sports':0.1}
collection = [doc1, doc2, doc3, doc4]
df = pd.DataFrame(collection)
# Fill missing values with zeros
df.fillna(0, inplace=True)
# Get Feature Vectors
feature_matrix = df.as_matrix()
# Get cosine similarity (i.e. 1 - cosine_distance) between pairs
sims = 1-pairwise_distances(feature_matrix, metric='cosine')
# Get the ranking of the documents given document 0
# from most similar to least similar. Don't take into account
# the first document, because it will be the same that the query was about
ranking_1 = np.argsort(sims[0,:])[::-1][1:2]
print ranking_1
ranking_2 = np.argsort(sims[1,:])[::-1][1:2]
print ranking_2
This takes 4 documents with $N=7$ unique topics, fills missing values with zeros and creates a similarity matrix between all documents. Then querying for documents 1(Science=0.7) and 2(Art:0.5) the most similar other document in the collection, we surely get documents 3(Science:0.8) and 4(Art:0.6) correspondingly.
You can try more sophisticated approaches regarding clustering and other distance metrics.