metadata
dataset_info:
features:
- name: name
dtype: string
- name: description
dtype: string
- name: solutions
dtype: string
splits:
- name: train
num_bytes: 22782308
num_examples: 8139
download_size: 10785367
dataset_size: 22782308
configs:
- config_name: default
data_files:
- split: train
path: data/train-*
このデータセットは、deepmind/code_contestsからAtCoderのデータのみを抽出し、Pythonで正解として提出された最初のコードを取得して格納したものです。これにより、教師あり学習に適した形に整えられています。
This dataset is created by extracting only the data from AtCoder within deepmind/code_contests and retrieving the first Python code submission marked as correct. As a result, it has been formatted to be suitable for supervised learning.
Example
description
There are n persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends.
We want to plan a trip for every evening of m days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold:
* Either this person does not go on the trip,
* Or at least k of his friends also go on the trip.
Note that the friendship is not transitive. That is, if a and b are friends and b and c are friends, it does not necessarily imply that a and c are friends.
For each day, find the maximum number of people that can go on the trip on that day.
Input
The first line contains three integers n, m, and k (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ m ≤ 2 ⋅ 10^5, 1 ≤ k < n) — the number of people, the number of days and the number of friends each person on the trip should have in the group.
The i-th (1 ≤ i ≤ m) of the next m lines contains two integers x and y (1≤ x, y≤ n, x≠ y), meaning that persons x and y become friends on the morning of day i. It is guaranteed that x and y were not friends before.
Output
Print exactly m lines, where the i-th of them (1≤ i≤ m) contains the maximum number of people that can go on the trip on the evening of the day i.
Examples
Input
4 4 2
2 3
1 2
1 3
1 4
Output
0
0
3
3
Input
5 8 2
2 1
4 2
5 4
5 2
4 3
5 1
4 1
3 2
Output
0
0
0
3
3
4
4
5
Input
5 7 2
1 5
3 2
2 5
3 4
1 2
5 3
1 3
Output
0
0
0
0
3
4
4
Note
In the first example,
* 1,2,3 can go on day 3 and 4.
In the second example,
* 2,4,5 can go on day 4 and 5.
* 1,2,4,5 can go on day 6 and 7.
* 1,2,3,4,5 can go on day 8.
In the third example,
* 1,2,5 can go on day 5.
* 1,2,3,5 can go on day 6 and 7.
solutions
#```python
from collections import deque
def solve(adj, m, k, uv):
n = len(adj)
nn = [len(a) for a in adj]
q = deque()
for i in range(n):
if nn[i] < k:
q.append(i)
while q:
v = q.popleft()
for u in adj[v]:
nn[u] -= 1
if nn[u] == k-1:
q.append(u)
res = [0]*m
nk = len([1 for i in nn if i >= k])
res[-1] = nk
for i in range(m-1, 0, -1):
u1, v1 = uv[i]
if nn[u1] < k or nn[v1] < k:
res[i - 1] = nk
continue
if nn[u1] == k:
q.append(u1)
nn[u1] -= 1
if not q and nn[v1] == k:
q.append(v1)
nn[v1] -= 1
if not q:
nn[u1] -= 1
nn[v1] -= 1
adj[u1].remove(v1)
adj[v1].remove(u1)
while q:
v = q.popleft()
nk -= 1
for u in adj[v]:
nn[u] -= 1
if nn[u] == k - 1:
q.append(u)
res[i - 1] = nk
return res
n, m, k = map(int, input().split())
a = [set() for i in range(n)]
uv = []
for i in range(m):
u, v = map(int, input().split())
a[u - 1].add(v - 1)
a[v - 1].add(u - 1)
uv.append((u-1, v-1))
res = solve(a, m, k, uv)
print(str(res)[1:-1].replace(' ', '').replace(',', '\n'))
#```