AtCoder abc143

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 : PyCharm

import 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.stdin
    n = int(fp.readline())
    a = list(map(int, fp.readline().split()))
    a.sort()
    ans = 0
    for i in range(n - 2):
        j = i + 1
        k = j + 1
        while j < n - 1:
            while k < n:
                if a[i] + a[j] <= a[k]:
                    break
                k += 1
            if k == j + 1 and a[i] + a[j] <= a[k]:
                pass
            else:
                ans += k - j - 1
            j += 1
    print(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 : PyCharm

import 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.stdin
    n, 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 - 1
        dist[a][b] = dist[b][a] = c

    for 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] = 1

    for 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 - 1
        if 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 : PyCharm

import 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.stdin
    n = 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]] += 1
        b[cnt[a[i]]] += 1
    for i in range(1, n + 1):
        s[i] = b[i] + s[i - 1]

    for k in range(1, n + 1):
        hi, lo = n + 1, 1
        while lo < hi:
            mid = (lo + hi) >> 1
            if mid * k <= s[mid]:
                lo = mid + 1
            else:
                hi = mid
        print(lo - 1)


if __name__ == "__main__":
    main()