AtCoder ABC188

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文章讲述了在Atcoder编程竞赛中，作者分享了两个问题的解题策略，涉及模拟方法、数组排序、边界计算以及动态规划在求解具有拓扑性质和数学问题上的应用。

本期题比上期更简单，只有一个绿题，而且并不是很难的数学题，估计比赛时能AK

C - ABC Tournament

模拟，记录一下位置

# -*- coding: utf-8 -*-
# @time     : 2023/6/2 13:30
# @file     : atcoder.py
# @software : PyCharm

import bisect
import copy
import sys
from itertools import permutations
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()
    fp = open("in.txt") if items[0] == "3.10.6" else sys.stdin
    n = int(fp.readline())
    a = list(map(int, fp.readline().split()))
    ans = -1
    lt = [i for i in range(len(a))]
    for i in range(n):
        t = []
        m = len(lt)
        for j in range(m // 2):
            i0, i1 = j * 2, j * 2 + 1
            if a[lt[i0]] > a[lt[i1]]:
                t.append(lt[i0])
            else:
                t.append(lt[i1])
        if len(lt) == 2:
            ans = lt[0] if a[lt[0]] < a[lt[1]] else lt[1]
        lt = t
    print(ans + 1)


if __name__ == "__main__":
    main()

D - Snuke Prime

模拟题，我的思路是边界和中间的区间单独计算

# -******- coding: utf-8  -******-
# @time     : 2023/12/28 20:10
# @author   : c.d.
# @desc     :
# @file     : atcoder.py
# @software : PyCharm

import sys
from collections import defaultdict


def main():
    items = sys.version.split()
    fp = open("in.txt") if items[0] == "3.10.9" else sys.stdin
    n, m = map(int, fp.readline().split())
    arr = defaultdict(list)
    for i in range(n):
        a, b, c = map(int, fp.readline().split())
        arr[a].append((0, c))
        arr[b].append((1, c))
    keys = list(arr.keys())
    keys.sort()
    cur = 0
    ans = 0
    last = -1
    for k in keys:
        if last != -1 and k - last > 1:
            ans += min(cur, m) * (k - last - 1)
        drop = 0
        for op, v in arr[k]:
            if op == 0:
                cur += v
            else:
                drop += v
        ans += min(cur, m)
        cur -= drop
        last = k
    print(ans)


if __name__ == "__main__":
    main()

E - Peddler

有拓扑性质的题目，容易想到DP
记录一个在该点买入后的最大值和卖出后的最大值，
答案是遍历卖出后的最大值。

#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <string>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <algorithm>
#define LT(x) (x * 2)
#define RT(x) (x * 2 + 1)

using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi;


int n, m;
LL a[200020];
vi g[200020];
LL dp[200020][2];


int main() { 
    //freopen("in.txt", "r", stdin);
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; ++i)
        scanf("%lld", a + i);
    for (int i = 0; i < m; ++i) {
        int u, v;
        scanf("%d%d", &u, &v);
        u--, v--;
        g[v].push_back(u);
    }
    for (int i = 0; i < n; ++i)
        dp[i][1] = INT_MIN;
    LL ans = INT_MIN;
    for (int i = 0; i < n; ++i) {
        dp[i][0] = -a[i];
        for (int v : g[i]) {
            if (dp[v][0] + a[i] > dp[i][1]) {
                dp[i][1] = dp[v][0] + a[i];
            }
            if (dp[v][0] > dp[i][0])
                dp[i][0] = dp[v][0];
        }
        ans = max(ans, dp[i][1]);
    }
    printf("%lld\n", ans);
    return 0;
}

F - +1-1x2

数学题
正难则反
从y倒推x，详情见代码

#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <string>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <algorithm>
#define LT(x) (x * 2)
#define RT(x) (x * 2 + 1)

using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi;


LL x, y;
map<LL, LL> cnt;


LL get(LL a) {
    if (cnt.count(a))
        return cnt[a];
    if (a <= x)
        return x - a;
    LL ret = a - x;
    if (a & 1) {
        ret = min(ret, get((a + 1) / 2) + 2);
        ret = min(ret, get((a - 1) / 2) + 2);
    }
    else {
        ret = min(ret, get(a / 2) + 1);
    }
    return cnt[a] = ret;
}


int main() { 
    //freopen("in.txt", "r", stdin);
    scanf("%lld%lld", &x, &y);
    if (x >= y) {
        printf("%lld\n", x - y);
    }
    else {
        LL ans = get(y);
        printf("%lld\n", ans);
    }
    return 0;
}