D - Triangles
排序后two pointer

# -*- coding: utf-8 -*-
# @time     : 2023/6/2 13:30
# @author   : yhdu@tongwoo.cn
# @desc     :
# @file     : atcoder.py
# @software : PyCharmimport bisect
import copy
import sys
from sortedcontainers import SortedList
from collections import defaultdict, Counter, deque
from functools import lru_cache, cmp_to_key
import heapq
import math
sys.setrecursionlimit(50050)def main():items = sys.version.split()if items[0] == '3.10.6':fp = open("in.txt")else:fp = sys.stdinn = int(fp.readline())a = list(map(int, fp.readline().split()))a.sort()ans = 0for i in range(n - 2):j = i + 1k = j + 1while j < n - 1:while k < n:if a[i] + a[j] <= a[k]:breakk += 1if k == j + 1 and a[i] + a[j] <= a[k]:passelse:ans += k - j - 1j += 1print(ans)if __name__ == "__main__":main()

E - Travel by Car
稍微有点绕
单纯floyd求最短路径是不行的，因为如一条路径是2,3,2，从点1到点3最短路径是7，而l=4时从点1到点4要加两次油。第二个图中如果路径为2,2，3, 最短路径还是7，但是只需要加一次油。


问题的关键是，第一个图中点1到点3是无法一次到达，而图2中可以。
因此修改原最短路径图为
如果点s到点t可以一次到达，点s到点t的距离设为1，否则设为INF
再求一次最短路径就是答案

# -*- coding: utf-8 -*-
# @time     : 2023/6/2 13:30
# @author   : yhdu@tongwoo.cn
# @desc     :
# @file     : atcoder.py
# @software : PyCharmimport bisect
import copy
import sys
from sortedcontainers import SortedList
from collections import defaultdict, Counter, deque
from functools import lru_cache, cmp_to_key
import heapq
import math
sys.setrecursionlimit(50050)def main():items = sys.version.split()if items[0] == '3.10.6':fp = open("in.txt")else:fp = sys.stdinn, m, l = map(int, fp.readline().split())dist = [[10 ** 18] * n for _ in range(n)]for i in range(m):a, b, c = map(int, fp.readline().split())a, b = a - 1, b - 1dist[a][b] = dist[b][a] = cfor i in range(n):for j in range(n):for k in range(n):if dist[j][i] + dist[i][k] < dist[j][k]:dist[j][k] = dist[i][k] + dist[j][i]cnt = [[10 ** 18] * n for _ in range(n)]for i in range(n):for j in range(n):if dist[i][j] <= l:cnt[i][j] = 1for i in range(n):for j in range(n):for k in range(n):if cnt[j][i] + cnt[i][k] < cnt[j][k]:cnt[j][k] = cnt[j][i] + cnt[i][k]q = int(fp.readline())for i in range(q):s, t = map(int, fp.readline().split())s, t = s - 1, t - 1if cnt[s][t] >= 10 ** 18:print(-1)else:print(cnt[s][t] - 1)if __name__ == "__main__":main()

F - Distinct Numbers

本题需要证明单调性质
待写

# -*- coding: utf-8 -*-
# @time     : 2023/6/2 13:30
# @author   : yhdu@tongwoo.cn
# @desc     :
# @file     : atcoder.py
# @software : PyCharmimport bisect
import copy
import sys
from sortedcontainers import SortedList
from collections import defaultdict, Counter, deque
from functools import lru_cache, cmp_to_key
import heapq
import math
sys.setrecursionlimit(50050)def main():items = sys.version.split()if items[0] == '3.10.6':fp = open("in.txt")else:fp = sys.stdinn = int(fp.readline())a = list(map(int, fp.readline().split()))cnt = [0] * (n + 1)b = [0] * (n + 1)s = [0] * (n + 1)for i in range(n):cnt[a[i]] += 1b[cnt[a[i]]] += 1for i in range(1, n + 1):s[i] = b[i] + s[i - 1]for k in range(1, n + 1):hi, lo = n + 1, 1while lo < hi:mid = (lo + hi) >> 1if mid * k <= s[mid]:lo = mid + 1else:hi = midprint(lo - 1)if __name__ == "__main__":main()