CSES - Aalto Competitive Programming 2024 - wk2 - Homework - Results
Submission details
Task:Connect cities
Sender:clovis
Submission time:2024-09-08 11:49:32 +0300
Language:Python3 (CPython3)
Status:READY
Result:
Test results
testverdicttime
#1ACCEPTED0.02 sdetails
#2ACCEPTED0.02 sdetails
#3ACCEPTED0.02 sdetails
#4ACCEPTED0.02 sdetails
#5ACCEPTED0.02 sdetails
#6ACCEPTED0.99 sdetails
#7ACCEPTED0.99 sdetails
#8--details
#9ACCEPTED0.99 sdetails
#10ACCEPTED0.97 sdetails
#11ACCEPTED0.29 sdetails
#12ACCEPTED0.02 sdetails

Code

def findRoadsToBuild(cityMappingDict, n):
  graphs = []
  visited = set()

  for city in range(1, n + 1):
    if city not in visited:
      graph = []
      dfs(city, cityMappingDict, visited, graph)
      graphs.append(graph)

  roadsToBuildList = []
  for i in range(1, len(graphs)):
    roadsToBuildList.append((graphs[i - 1][0], graphs[i][0]))

  return roadsToBuildList

def dfs(city, cityMappingDict, visited, graph):
  stack = [city]
  visited.add(city)

  while stack:
    currCity = stack.pop()
    graph.append(currCity)

    for neighbour in cityMappingDict[currCity]:
      if neighbour not in visited:
        visited.add(neighbour)
        stack.append(neighbour)

# first line, n (number of cities) and m (number of roads)
numCitiesNumRoadsInput = input().split()
n = int(numCitiesNumRoadsInput[0])
m = int(numCitiesNumRoadsInput[1])

if n == 1:
  print(0)
elif n == 2 and m == 1:
  print(0)
else:
  # initialise dict to act as an adjacency list
  cityMappingDict = dict()

  # subsequent lines, roads connecting cities
  for i in range(m):
    cityRoadConnectionString = input().split()
    leftCity = int(cityRoadConnectionString[0])
    rightCity = int(cityRoadConnectionString[1])

    if leftCity not in cityMappingDict:
      cityMappingDict[leftCity] = []
    if rightCity not in cityMappingDict:
      cityMappingDict[rightCity] = []

    cityMappingDict[leftCity].append(rightCity)
    cityMappingDict[rightCity].append(leftCity)

  # fill in missing cities who don't have any roads
  for i in range(1, n + 1):
    if i not in cityMappingDict:
      cityMappingDict[i] = []

  # print(cityMappingDict)

  newRoads = findRoadsToBuild(cityMappingDict, n)

  print(len(newRoads))

  for newRoad in newRoads:
    print(newRoad[0], newRoad[1])

Test details

Test 1

Verdict: ACCEPTED

input
10 10
2 5
5 6
1 4
6 8
...

correct output
2
1 2
2 7

user output
2
1 2
2 7

Test 2

Verdict: ACCEPTED

input
10 10
3 9
6 8
9 10
7 8
...

correct output
2
1 4
4 5

user output
2
1 4
4 5

Test 3

Verdict: ACCEPTED

input
10 10
7 9
1 7
1 3
3 4
...

correct output
0

user output
0

Test 4

Verdict: ACCEPTED

input
10 10
4 8
5 9
4 9
2 7
...

correct output
1
1 3

user output
1
1 3

Test 5

Verdict: ACCEPTED

input
10 10
4 9
2 4
7 10
1 8
...

correct output
0

user output
0

Test 6

Verdict: ACCEPTED

input
100000 200000
7233 22146
94937 96203
6133 10731
98737 99193
...

correct output
4785
1 2
2 3
3 4
4 5
...

user output
4785
1 2
2 3
3 4
4 5
...
Truncated

Test 7

Verdict: ACCEPTED

input
100000 200000
92950 93575
24401 88897
41796 99364
47106 50330
...

correct output
4868
1 2
2 7
7 9
9 15
...

user output
4868
1 2
2 7
7 9
9 15
...
Truncated

Test 8

Verdict:

input
100000 200000
15637 76736
79169 98809
4382 86557
73383 77029
...

correct output
4683
1 9
9 20
20 27
27 28
...

user output
(empty)

Test 9

Verdict: ACCEPTED

input
100000 200000
47932 66981
86401 99942
4353 27841
60492 67345
...

correct output
4807
1 6
6 7
7 11
11 12
...

user output
4807
1 6
6 7
7 11
11 12
...
Truncated

Test 10

Verdict: ACCEPTED

input
100000 200000
6554 44548
76413 98555
5447 59589
70166 74434
...

correct output
4786
1 2
2 18
18 21
21 27
...

user output
4786
1 2
2 18
18 21
21 27
...
Truncated

Test 11

Verdict: ACCEPTED

input
100000 1
1 2

correct output
99998
1 3
3 4
4 5
5 6
...

user output
99998
1 3
3 4
4 5
5 6
...
Truncated

Test 12

Verdict: ACCEPTED

input
10 9
2 5
5 6
1 4
6 8
...

correct output
2
1 2
2 7

user output
2
1 2
2 7