【codeforces 348D】Turtles

本文介绍LGV定理在二维空间中计算从起点集到终点集不相交路径方案数的应用,通过矩阵的行列式计算路径方案数,并提出使用暴力dp方法求解。

题目描述

社论

LGV定理告诉我们,在二维空间下,有起点集 $S$,以及终点集 $T$,从 $S$ 到 $T$ 的每一个点走到对应点且不相交路径的方案数为下列矩阵的行列式:

$$
\begin{pmatrix}
f(S_1,T_1) & f(S_1,T_2) & f(S_1,T_3) & \cdots & f(S_1,T_n) \\
f(S_2,T_1) & f(S_2,T_2) & f(S_2,T_3) & \cdots & f(S_2,T_n) \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
f(S_n,T_1) & f(S_n,T_2) & f(S_n,T_3) & \cdots & f(S_n,T_n)
\end{pmatrix}
$$

其中 $f(S,T)$ 表示从 $S$ 到 $T$ 的方案数

然后就每次暴力 $dp$ 一下即可

转载于:https://www.cnblogs.com/KingSann/articles/11128552.html

源码地址: https://pan.quark.cn/s/d1f41682e390 miyoubiAuto 米游社每日米游币自动化Python脚本(务必使用Python3) 8更新:更换cookie的获取地址 注意:禁止在B站、贴吧、或各大论坛大肆传播! 作者已退游,项目不维护了。 如果有能力的可以pr修复。 小引一波 推荐关注几个非常可爱有趣的女孩! 欢迎B站搜索: @嘉然今天吃什么 @向晚大魔王 @乃琳Queen @贝拉kira 第三方库 食用方法 下载源码 在Global.py中设置米游社Cookie 运行myb.py 本地第一次运行时会自动生产一个文件储存cookie,请勿删除 当前仅支持单个账号! 获取Cookie方法 浏览器无痕模式打开 http://user.mihoyo.com/ ,登录账号 按,打开,找到并点击 按刷新页面,按下图复制 Cookie: How to get mys cookie 当触发时,可尝试按关闭,然后再次刷新页面,最后复制 Cookie。 也可以使用另一种方法: 复制代码 浏览器无痕模式打开 http://user.mihoyo.com/ ,登录账号 按,打开,找到并点击 控制台粘贴代码并运行,获得类似的输出信息 部分即为所需复制的 Cookie,点击确定复制 部署方法--腾讯云函数版(推荐! ) 下载项目源码和压缩包 进入项目文件夹打开命令行执行以下命令 xxxxxxx为通过上面方式或取得米游社cookie 一定要用双引号包裹!! 例如: png 复制返回内容(包括括号) 例如: QQ截图20210505031552.png 登录腾讯云函数官网 选择函数服务-新建-自定义创建 函数名称随意-地区随意-运行环境Python3....
### Codeforces 1487D Problem Solution The problem described involves determining the maximum amount of a product that can be created from given quantities of ingredients under an idealized production process. For this specific case on Codeforces with problem number 1487D, while direct details about this exact question are not provided here, similar problems often involve resource allocation or limiting reagent type calculations. For instance, when faced with such constraints-based questions where multiple resources contribute to producing one unit of output but at different ratios, finding the bottleneck becomes crucial. In another context related to crafting items using various materials, it was determined that the formula `min(a[0],a[1],a[2]/2,a[3]/7,a[4]/4)` could represent how these limits interact[^1]. However, applying this directly without knowing specifics like what each array element represents in relation to the actual requirements for creating "philosophical stones" as mentioned would require adjustments based upon the precise conditions outlined within 1487D itself. To solve or discuss solutions effectively regarding Codeforces' challenge numbered 1487D: - Carefully read through all aspects presented by the contest organizers. - Identify which ingredient or component acts as the primary constraint towards achieving full capacity utilization. - Implement logic reflecting those relationships accurately; typically involving loops, conditionals, and possibly dynamic programming depending on complexity level required beyond simple minimum value determination across adjusted inputs. ```cpp #include <iostream> #include <vector> using namespace std; int main() { int n; cin >> n; vector<long long> a(n); for(int i=0;i<n;++i){ cin>>a[i]; } // Assuming indices correspond appropriately per problem statement's ratio requirement cout << min({a[0], a[1], a[2]/2LL, a[3]/7LL, a[4]/4LL}) << endl; } ``` --related questions-- 1. How does identifying bottlenecks help optimize algorithms solving constrained optimization problems? 2. What strategies should contestants adopt when translating mathematical formulas into code during competitive coding events? 3. Can you explain why understanding input-output relations is critical before implementing any algorithmic approach? 4. In what ways do prefix-suffix-middle frameworks enhance model training efficiency outside of just tokenization improvements? 5. Why might adjusting sample proportions specifically benefit models designed for tasks requiring both strong linguistic comprehension alongside logical reasoning skills?
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值