本文版权归ljh2000和博客园共有,欢迎转载,但须保留此声明,并给出原文链接,谢谢合作. 本文作者:ljh2000 作者博客:http://www.cnblogs.com/ljh2000-jump/转载请注明出处,侵权必究,保留最终解释权! 题目链接:CF27D 正解:$2—SAT$ 解题报告: 也是$2-SAT$裸题,判断线段相交之后,直接上$2-SAT$就好了. 但是判线段相交有一点难受啊,换了几种写法才搞对,最好是双向判断一下…' 注意输出方案的时候一般的做法就是将图反向之后拓扑排序,一…

A. Ring road time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Nowadays the one-way traffic is introduced all over the world in order to improve driving safety and reduce traffic jams. The g…

    题意             n个数1~n按顺序围成一个圈...现在在某些两点间加边..边可以加在圈内或者圈外..问是否会发生冲突?如果不发生冲突..输每一条边是放圈内还是圈外.     题解             这道题和POJ 3207差不多了..只是那道题只要判断是否存在不要输出方案...发现个很严重的问题..POJ 3207的数据实在是太弱了..我上一个程序里判断两个线段是否相交是个错了..都让我AC了..导致我做这题是沿用了思路...浪费了很多时间...          …

题目大意 给你一个\(n\)个点的环,每条边有方向,改变第\(i\)条边的方向代价为\(w_i\),问将其改为强连通图的最小代价.\(n\leqslant100\) 题解 求出把边全部改为顺时针和全部改为逆时针的代价,较少的输出即可 卡点 无 C++ Code: #include <cstdio> #include <iostream> #include <algorithm> const int maxn = 111; int head[maxn], cnt; str…

In this lesson you will learn to answer simple questions about yourself.  本节课讲学到回答关于自己的一些简单问题 课上内容(Lesson) Hello -->  Hi Thank you -->  Thanks I'm fine -->  I'm OK My name is ... -->  I'm # Meeting People What's you nickname? You can call me .…

For success, attitude is equally as important as ability. 为取得成功,态度与能力一样重要. Today I read a news about the world's first photovolatic highway which is currently built in Jinan, the construction work nearly comes to the end. According to the news, the 2…

146. The Runner time limit per test: 0.25 sec.memory limit per test: 4096 KB input: standard inputoutput: standard output The runner moves along the ring road with length L. His way consists of N intervals. First he ran T1 minutes with speed V1, then…

目录 CSP2019-S游记 Day -2(UPDATE:2019-11-14) Day -1(UPDATE:2019-11-15) Day 1(UPDATE:2019-11-16) Day 2(UPDATE:2019-11-17) Day3(UPDATE:2019-11-18) CSP2019-S游记 Day -2(UPDATE:2019-11-14) 完了,我写游记了,是不是要被禁赛了. 刚刚写完不一定会有游记,然后开始写,感觉自己打脸真/cy快. 经过\(11,12,13\)连续三天的期中…

目录 CSP2019-S宝典 模板 博客 快读 vim配置 对拍 CSP2019-S注意事项 考前 考时 考后 游记 Day -2(UPDATE:2019-11-14) Day -1(UPDATE:2019-11-15) Day 1(UPDATE:2019-11-16) Day 2(UPDATE:2019-11-17) Day3(UPDATE:2019-11-18) CSP2019-S宝典 --\(Memory\_of\_winter\) 放图镇楼 模板 博客 博客模板链接 如平衡树,网络流,最…

前言 首先先刷完这些在说 题单 25C Roads in Berland 25D Roads not only in Berland 9E Interestring graph and Apples 14D Two Paths 20C Dijkstra? 22E Scheme ※Mark {27D Ring Road 2}:2-SAT,都这么强的吗?什么都会? (学了2-SAT再来做) 29E Quarrel 卡我时间 33D Knights ※Mark {46F Hercule Poirot…

一个队列如果只生产不消费肯定不行的,那么如何及时消费Ring Buffer的数据呢?简单的方案就是当Ring Buffer"写满"的时候一次性将数据"消费"掉.注意这里的"写满"仅仅是指写入位置 index达到了数组最大索引位置,而"消费"也不同于常见的堆栈,队列等数据结构,只是读取缓冲区的数据而不会移除它.一点公益系统 一点公益系统模式 一点公益平台 一点公益APP系统 一点公益商业模式定制找陈牧150-1315-1740(…

OpenCASCADE Ring Type Spring Modeling eryar@163.com Abstract. The general method to directly create an edge is to give a 3D curve as the support(geometric domain) of the edge. The curve maybe defined as a 2D curve in the parametric space of a surface…

最近常收到SOD框架的朋友报告的SOD的SQL日志功能报错:文件句柄丢失.经过分析得知,这些朋友使用SOD框架开发了访问量比较大的系统,由于忘记关闭SQL日志功能所以出现了很高频率的日志写入操作,从而偶然引起错误.后来我建议只记录出错的或者执行时间较长的SQL信息,暂时解决了此问题.但是作为一个热心造轮子的人,一定要看看能不能造一个更好的轮子出来. 前面说的错误原因已经很直白了,就是频繁的日志写入导致的,那么解决方案就是将多次写入操作合并成一次写入操作,并且采用异步写入方式.要保存多次操作的内容…

AlwaysOn Ring Buffers 一些AlwaysOn的诊断信息可以从SQL Server ring buffers.或者从sys.dm_os_ring_buffers.ring buffer在SQL Server启动的时候被创建,SQL Server系统为内部诊断记录警告. 通过以下查询获取所有事件记录 SELECT * FROM sys.dm_os_ring_buffers WHERE ring_buffer_type LIKE '%HADR%' 为了让数据更加可控,可以通过日期,…

http://codeforces.com/contest/738/problem/C Vasya is currently at a car rental service, and he wants to reach cinema. The film he has bought a ticket for starts in t minutes. There is a straight road of length s from the service to the cinema. Let's…

Ikki's Story I - Road Reconstruction Time Limit: 2000MS   Memory Limit: 131072K Total Submissions: 7659   Accepted: 2215 Description Ikki is the king of a small country – Phoenix, Phoenix is so small that there is only one city that is responsible fo…

传送门 Description A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 1, 2, . . . , n into each circle separately, and the sum of numbers in two adjacent circles should be a prime. Note: the number of first circle sho…

Ring buffers and queues The data structure is extremely simple: a bounded FIFO. One step up from plain arrays, but still, it’s very basic stuff. And if you’re doing system programming, particularly anything involving IO or directly talking to hardwar…

Problem Description A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime. Note: the number of first circle should always be 1…

Problem Description A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime. Note: the number of first circle should always be 1…

C. Road to Cinema time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vasya is currently at a car rental service, and he wants to reach cinema. The film he has bought a ticket for starts in t m…

Road Trip Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KB Total submit users: 29, Accepted users: 29 Problem 12882 : No special judgement Problem description You are planning a road trip to visit your friends, each of whom live in…

题目链接: 传送门 find the most comfortable road Time Limit: 1000MS     Memory Limit: 32768 K Description XX星有许多城市,城市之间通过一种奇怪的高速公路SARS(Super Air Roam Structure---超级空中漫游结构)进行交流,每条SARS都对行驶在上面的Flycar限制了固定的Speed,同时XX星人对 Flycar的"舒适度"有特殊要求,即乘坐过程中最高速度与最低速度的差越小…

Design road Problem's Link:   http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1548 Mean: 目的:从(0,0)到达(x,y).但是在0~x之间有n条平行于y轴的河,每条河位于xi处,无限长,wi宽,并分别给出了建立路和桥每公里的单价 求:到达目标的最小费用. analyse: 比赛的时候一直没想到思路,第二个样列怎么算都算不对,赛后才知道是三分. 首先把所有的桥移到最右端,然后三分枚举路和河的交点. Time…

  Prime Ring Problem  A ring is composed of n (even number) circles as shown in diagram. Put natural numbers into each circle separately, and the sum of numbers in two adjacent circles should be a prime. Note: the number of first circle should always…

传送门:hdu 5861 Road 题意: 水平线上n个村子间有 n-1 条路. 每条路开放一天的价格为 Wi 有 m 天的操作,每天需要用到村子 Ai~Bi 间的道路 每条路只能开放或关闭一次. (不能重复开关) 求每天的最小花费. 思路: 第一次线段树:维护每条路第一次和最后一次被用到的天数.以下代码维护了 mn:第一次被用到,mx:最后一次被用到,lazy:被更新的最大值若当前区间被lazy维护而没有更新到点,那么这个子节点的最小值就可能被改变.所以我这里的子节点更新是根据父节点的最大和最…

Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 18313    Accepted Submission(s): 8197 Problem Description A ring is compose of n circles as shown in diagram. Put natural numbe…

枚举作为magic road的边,然后求出A/B. A/B得在大概O(1)的时间复杂度求出,关键是B,B是包含magic road的最小生成树. 这么求得: 先在原图求MST,边总和记为s,顺便求出MST上任意两点路径上的最长边d[i][j]. 当(u,v)是magic road时, 如果它在原本的MST上,则B就等于s-原(u,v)的权,而原(u,v)的权其实就是d[u][v]: 如果它不在原本的MST上,则B就等于s-d[u][v]+0. 总之就是一个式子:B=s-d[u][v]. 于是,在…

Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 42600    Accepted Submission(s): 18885 Problem Description A ring is compose of n circles as shown in diagram. Put natural num…

Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 23458    Accepted Submission(s): 10465 Problem Description A ring is compose of n circles as shown in diagram. Put natural numb…