AtCoder ABC132 - 技术栈

陷入了"会做，但不完全会做"的状况

水题，排序找中间数两边的差值

组合数学

求把n个相同的球分到m个相同的盒子，1每个盒子至少一个球2每个盒子球不限的组合数

空挡插隔板法，高中数学

python 复制代码

# -*- coding: utf-8 -*-
# @time     : 2023/6/2 13:30
# @author   : [email protected]
# @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(100010)

mod = 10 ** 9 + 7


def pw(a, x):
    if x == 0:
        return 1
    temp = pw(a, x >> 1)
    if x & 1:
        return temp * temp * a % mod
    else:
        return temp * temp % mod


def inv(x):
    return pw(x, mod - 2)


def main():
    items = sys.version.split()
    if items[0] == '3.10.6':
        fp = open("in.txt")
    else:
        fp = sys.stdin
    n, k = map(int, fp.readline().split())
    R, B = n - k, k

    def cmb(x, y):
        if x < y:
            return 0
        return fac[x] * inv(fac[y]) * inv(fac[x - y]) % mod

    fac = [1] * 5005
    for i in range(2, 5005):
        fac[i] = fac[i - 1] * i % mod

    for i in range(1, B + 1):
        ans_r = cmb(R + 1, i)
        ans = cmb(B - 1, i - 1) * ans_r % mod
        print(ans)


if __name__ == "__main__":
    main()

一开始用dfs去求（3步能访问到的点对），明显TLE

分层图求最短路

原图分为3层，每个点拆成 V 0 , V 1 , V 2 V_0,V_1,V_2 V0,V1,V2，

假如 ( u , v ) (u,v) (u,v)间有一条有向边

把边拆成 ( u 0 , v 1 ) , ( u 1 , v 2 ) , ( u 2 , v 3 ) (u_0,v_1),(u_1,v_2),(u_2,v_3) (u0,v1),(u1,v2),(u2,v3)

走三的倍数步能回到0层，这样只要能走到 T 0 T_0 T0点就有解

python 复制代码

# -*- coding: utf-8 -*-
# @time     : 2023/6/2 13:30
# @author   : [email protected]
# @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(100010)


def main():
    items = sys.version.split()
    if items[0] == '3.10.6':
        fp = open("in.txt")
    else:
        fp = sys.stdin
    n, m = map(int, fp.readline().split())
    g = [[] for _ in range(n * 3)]
    for i in range(m):
        u, v = map(int, fp.readline().split())
        u, v = u - 1, v - 1
        u0, u1, u2 = u, u + n, u + 2 * n
        v0, v1, v2 = v, v + n, v + 2 * n
        g[u0].append(v1)
        g[u1].append(v2)
        g[u2].append(v0)
    S, T = map(int, fp.readline().split())
    S, T = S - 1, T - 1

    vis = [-1] * (3 * n)
    qu = deque()
    qu.append(S)
    vis[S] = 0
    while qu:
        cur = qu.popleft()
        if cur == T:
            break
        for v in g[cur]:
            if vis[v] == -1:
                vis[v] = vis[cur] + 1
                qu.append(v)
    if vis[T] == -1:
        print(-1)
    else:
        print(vis[T] // 3)


if __name__ == "__main__":
    main()

分块

重要的是想清楚每一块代表的数字