AtCoder abc142

文章介绍了在ATCoder编程平台上遇到的三个问题,涉及位置数组遍历、计算gcd中质因数个数的dp算法,以及使用BFS求解纯最小环问题。

本期四道水题

C - Go to School
位置数组遍历

D - Disjoint Set of Common Divisors
简单的数论题
求gcd中的质因数个数+1

E - Get Everything
简单dp。会位操作就会做。

# -*- 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 = map(int, fp.readline().split())
    a = [0] * m
    c = [0] * m
    for i in range(m):
        t, k = map(int, fp.readline().split())
        a[i] = t
        b = list(map(int, fp.readline().split()))
        for j in range(k):
            c[i] ^= 1 << b[j] - 1

    dp = [[10 ** 18] * (1 << n) for i in range(m + 1)]
    dp[0][0] = 0
    for i in range(m):
        for j in range(1 << n):
            if dp[i][j] >= 10 ** 18:
                continue
            t, val = c[i - 1], a[i - 1]
            dp[i + 1][j | t] = min(dp[i][j] + val, dp[i + 1][j | t])
            dp[i + 1][j] = min(dp[i + 1][j], dp[i][j])
    ans = dp[m][(1 << n) - 1]
    if ans >= 10 ** 18:
        print(-1)
    else:
        print(ans)


if __name__ == "__main__":
    main()

F - Pure
求最小环,简单的bfs就好
什么tarjan的都弱爆了

# -*- 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 = map(int, fp.readline().split())
    g = [[] for _ in range(n)]
    for i in range(m):
        a, b = map(int, fp.readline().split())
        a -= 1
        b -= 1
        g[a].append(b)
    dist = [[10 ** 18] * n for i in range(n)]
    pre = [[-1] * n for i in range(n)]

    for i in range(n):
        vis = [0] * n
        qu = deque()
        qu.append(i)
        dist[i][i] = 0
        while qu:
            cur = qu.popleft()
            for v in g[cur]:
                if vis[v] == 0:
                    vis[v] = 1
                    qu.append(v)
                    dist[i][v] = dist[i][cur] + 1
                    pre[i][v] = cur

    ans = 10 ** 18
    sel = -1
    for i in range(n):
        if dist[i][i] != 0:
            if dist[i][i] < ans:
                ans, sel = dist[i][i], i
    if ans == 10 ** 18:
        print(-1)
    else:
        print(ans)
        cur = sel
        for i in range(ans):
            print(cur + 1)
            cur = pre[sel][cur]


if __name__ == "__main__":
    main()

由于无法直接获取AtCoder ABC比赛的最新题解,可通过AtCoder官方网站(https://atcoder.jp/)查看最新比赛场次及题目,比赛结束后,在社区论坛如AtCoder的Discuss板块、GitHub等地方能找到他人分享的题解。 参考之前不同场次题解示例,例如AtCoder ABC 171 C题,是将一个数N转化为26进制用abc表示,模拟过程需注意细节,其AC代码如下: ```cpp #include<iostream> #include<cstdio> #include <stdio.h> #include<algorithm> #include<cstring> #include<cmath> #include<cstdlib> #include<queue> #include<map> #include<vector> #include <set> #define ll long long using namespace std; char a[5000000]; int main() { ll n; ll c=1; int m=0; cin>>n; while(n>0) { ll k=n%(26); if(k==0) { k=26; a[m++]='z'; } else { a[m++]='a'+k-1; } n=(n-k)/26; } for(int i=m-1;i>=0;i--) { cout<<a[i]; } return 0; } ``` 又如AtCoder ABC183题的AC代码如下: ```cpp #include<iostream> #include<bit/stdc++.h> #include<algorithm> #include<vector> #include<numeric> using namespace std; int main(void) { ios_base::sync_with_stdio(false); int N,W; cin >> N >> W; vector<int64_t> v(300000); for(int i = 0; i < N; i++){ int S, T, P; cin >> S >> T >> P; v[S] += P; v[T] -= P; } partial_sum(v.begin(), v.end(), v.begin()); if(*max_element(v.begin(), v.end()) > W) { cout << "No" << endl; } else { cout << "Yes" << endl; } return 0; } ``` 再如AtCoder ABC 242题解,使用while循环往上跳,每次字符往后移动1或2,开变量记录,时间复杂度为$O(Q\log k)$,代码如下: ```cpp #include <bits/stdc++.h> using namespace std; typedef unsigned long long uLL; typedef long double LD; typedef long long LL; typedef double db; const int N = 100005; int n, Ti; char x[N]; int main() { scanf("%s%d", x + 1, &Ti); n = strlen(x + 1); for (LL a, b, c; Ti--; ) { scanf("%lld%lld", &a, &b); c = 0; while (a && b > 1) { if (!(b & 1)) ++c; ++c; b = b + 1 >> 1, a--; } c += a; printf("%c\n", char((x[b] - 'A' + c) % 3 + 'A')); } } ``` 还有AtCoder ABC237题是签到题,可使用C++的INT_MAX和INT_MIN,若不知这两个常数可自行定义,AC代码如下: ```cpp #include<bits/stdc++.h> using namespace std; typedef long long LL; int main(){ ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); LL n; cin>>n; if (n>=INT_MIN && n<=INT_MAX){ cout<<"Yes\n"; }else{ cout<<"No\n"; } return 0; } ```
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