Update README.md

README.md

@@ -23,3 +23,180 @@ configs:
 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'))
+```
+```