题意:给你一个无向图,如今要求你把边改成有向的. 使得入度为0的点最少,输出有多少个点入度为0 思路:脑补一波结论.假设有环的话显然没有点入度为0,其余则至少有一个点入度为0,然后就DFS一波就能够了 #include <cstdio> #include <queue> #include <cstring> #include <iostream> #include <cstdlib> #include <algorithm> #inc…

题目链接: http://codeforces.com/problemset/problem/659/E 题意: 给定n个点和m条双向边,将双向边改为单向边,问无法到达的顶点最少有多少个? 分析: 无法到达的话即入度为0. DFS判断每一个连通块中是否存在环,如果存在环,就能保证环中每个点的入度都大于等于1.否则,有头有尾,头的入度为0. 代码: #include<cstdio> #include<queue> #include<cstring> #include<…

Description Berland has n cities connected by m bidirectional roads. No road connects a city to itself, and each pair of cities is connected by no more than one road. It isnot guaranteed that you can get from any city to any other one, using only the…

New Reform time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Berland has n cities connected by m bidirectional roads. No road connects a city to itself, and each pair of cities is connected b…

题目链接 对于每一个联通块, 如果有一个强连通分量, 那么这个联通块对答案的贡献就是0. 否则对答案贡献是1. #include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <complex> #include <cmath> #include <map> #inc…

题目链接: E. New Reform time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Berland has n cities connected by m bidirectional roads. No road connects a city to itself, and each pair of cities is co…

题目链接:http://codeforces.com/problemset/problem/732/F 题意: 给出一个有n个点m条边的无向图,保证联通,现在要求将所有边给定一个方向使其变成有向图,设f(x)为点x能到达的点的个数,要求使最小的f(x)最大,并输出方案. 思路: tarjan一下,答案肯定是强连通分量里点最多的一个分量,而同一个强连通里的点成环,其他分量都指向这个最大点个数的分量. 退役了,偶尔刷一下题... #include <bits/stdc++.h> using nam…

边双连通分量. 这题有一点构造的味道.一个有向图,经过强连通缩点之后会形成一个有向无环图. 如果将最大的强连通分量放在顶端,其余的强连通分量都直接或间接指向他,那么这样就构造出了符合要求的图. 接下来就是要去寻找强连通分量.对于一个无向图来说,每一个边-双联通分量都可以将每条边定向之后构造成一个强连通分量,$dfs$一遍即可. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #…

构造. 有一种十分巧妙的方法可以使图中所有度数为偶数的节点,经过每条边定向后,出度和入度都相等. 首先统计每个节点的度数,将度数为奇数的节点与编号为$n+1$的节点连边,这样一来,这张新图变成了每个节点的度数都是偶数的图,必然存在欧拉回路.任意一条欧拉回路都使得原图中度数为偶数的节点的出度与入度相等. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cs…

E. One-Way Reform 题目连接: http://codeforces.com/contest/723/problem/E Description There are n cities and m two-way roads in Berland, each road connects two cities. It is known that there is no more than one road connecting each pair of cities, and ther…

E. New Reform 题目连接: http://www.codeforces.com/contest/659/problem/E Description Berland has n cities connected by m bidirectional roads. No road connects a city to itself, and each pair of cities is connected by no more than one road. It is not guara…

Description There are n cities and m two-way roads in Berland, each road connects two cities. It is known that there is no more than one road connecting each pair of cities, and there is no road which connects the city with itself. It is possible tha…

E. New Reform time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Berland has n cities connected by m bidirectional roads. No road connects a city to itself, and each pair of cities is connecte…

题目链接: 题目 E. New Reform time limit per test 1 second memory limit per test 256 megabytes inputstandard input outputstandard output 问题描述 Berland has n cities connected by m bidirectional roads. No road connects a city to itself, and each pair of cities…

E. New Reform time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Berland has n cities connected by m bidirectional roads. No road connects a city to itself, and each pair of cities is connecte…

[题目链接]:http://codeforces.com/contest/732/problem/F [题意] 给你一张无向图; n个点,m条边; 让你把这张图改成有向边 然后定义r[i]为每个点能够到达的其他点的数目; 让你使得最小的r[i]尽可能地大; 让你输出这个尽可能大的最小的ri; 然后输出改边之后的有向图; [题解] 如果整张图是一个环的话; 这个环上的每个点的答案就是确定的; 即为这个环的大小; 则考虑把原图缩点; 缩点之后; 每个环都能成为一个点; 则最后的答案就是强连通分量中所…

[题目链接]:http://codeforces.com/contest/723/problem/E [题意] 给你一个无向图; 让你把这m条边改成有向图; 然后使得出度数目等于入度数目的点的数目最多; 输出这个点的数目; 同时输出更改之后的所有有向边 [题解] 需要先明确; 就是图中度数为奇数的点的个数肯定是偶数个; 因为每条边都会贡献2度数; 所以最后的总度数肯定是偶数的; 则奇度数的点肯定得是偶数个,不然最后不会满足总度数为偶数; 知道这个之后; 猜想; 最后的答案就为原图中度数为偶数的点…

题目链接:传送门 题目大意:一副无向图,要求你给边定向(变为有向图),使出度等于入度的点最多,输出有多少 个点,并且输出定向后的边(前为起点,后为终点) 题目思路:欧拉路 我们这样考虑,先考虑无向图的点的度数,如果为奇数则一定无法变为题目要求的点,ans-1 对于度为偶数的点则一定可以通过调整满足. 处理方法:新建一个虚拟节点0,使所有度为奇数的点向其连一条边,那么最终图中的点的度数都为偶数. 这样就满足欧拉路的条件了.我们只需要跑欧拉路并且将走过的路径保留下来即可. 注意将与虚拟节点连的边删去…

前言:关于如何求双连通分量,我们可以在tarjan搜索时标记下所有桥的位置(双连通分量(可以认为是没有桥的无向图图)即可通过删去所有桥得到),那么怎么找桥呢,对于每一条搜索到的边u->x,如果low[u]>dfn[x]则说明u不能通过子图到达比x更早的节点,那么就说明该边是桥 题意:把一个无向图变成有向图,对于这个有向图来说,每个点的价值是它所有能到达的点的数量,要求使得所有点中最小的价值最大 题解:现学的边-双联通分量,先求一遍边-双联通分量,然后在每个双连通分量里dfs一边把双向边变成单向…

前面说点什么.. 为了完成日常积累,傻逼呵呵的我决定来一发codeforces 挑水题 泛做.. 嗯对,就是泛做.. 主要就是把codeforces Div.1的ABCD都尝试一下吧0.0.. 挖坑0.0.. Codeforces Round #360 A. NP-Hard Problem 就是一个二分图染色,判断是否为二分图而已.. B. Remainders Game 题意:有一个未知的\(x\)和已知\(x\ mod\ c_i\)的值,问是否能确定\(x\ mod\ k\)的值 考虑一下中…

Codeforces Round #346 (Div. 2)---E. New Reform E. New Reform time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Berland has n cities connected by m bidirectional roads. No road connects a city…

链接:https://codeforces.com/contest/659 A - Round House - [取模] AC代码: #include<bits/stdc++.h> using namespace std; int n,a,b; int main() { cin>>n>>a>>b; a=(a-+b); ) a+=n; a%=n; cout<<a+<<endl; } B - Qualifying Contest - [水…

题目链接:http://codeforces.com/contest/440/problem/D D. Berland Federalization   Recently, Berland faces federalization requests more and more often. The proponents propose to divide the country into separate states. Moreover, they demand that there is a…

Position:http://codeforces.com/contest/723 我的情况 啊哈哈,这次raiting肯定要涨,接受过上次的教训,先用小号送肉,大号都是一发切,重回蓝咯 结果... FST!! 不,这次是skip,A题gi了(从小号蒯来没改),其它就都会skip...... 大号被小号skip,大号还300多名(没WA),做不得卵声.结果小号rank408,+133raiting 恭喜LCFrank10,+279raiting 为了表示我被skip的愤怒,我决定用大号打Vit…

上一次我们拿学校的URP做了个小小的demo.... 其实我们还可以把每个学生的证件照爬下来做成一个证件照校花校草评比 另外也可以写一个物理实验自动选课... 但是出于多种原因,,还是绕开这些敏感话题.. 今天,我们来扒一下cf的题面! PS:本代码不是我原创 1. 必要的分析 1.1 页面的获取 一般情况CF的每一个 contest 是这样的: 对应的URL是:http://codeforces.com/contest/xxx 还有一个Complete problemset页面,它是这样的:…

http://codeforces.com/contest/738/problem/D Galya is playing one-dimensional Sea Battle on a 1 × n grid. In this game a ships are placed on the grid. Each of the ships consists of bconsecutive cells. No cell can be part of two ships, however, the shi…

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…

http://codeforces.com/contest/738/problem/A Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string s consisting of n lowercase English letters. There is…

http://codeforces.com/problemset/problem/662/A 题目大意: 给定n(n <= 500000)张卡片,每张卡片的两个面都写有数字,每个面都有0.5的概率是在正面,各个卡牌独立.求把所有卡牌来玩Nim游戏,先手必胜的概率. (⊙o⊙)-由于本人只会在word文档里写公式,所以本博客是图片格式的. Code #include <cstdio> #include <cstring> #include <algorithm> u…

http://codeforces.com/problemset/problem/274/B 题目大意: 给定你一颗树,每个点上有权值. 现在你每次取出这颗树的一颗子树(即点集和边集均是原图的子集的连通图)且这颗树中必须包含节点1 然后将这颗子树中的所有点的点权+1或-1 求把所有点权全部变为0的最小次数(n<=10^5) 题解: 因为每一次的子树中都必须有1,所以我们得知每一次变换的话1的权值都会变化 所以我们以1为根 现在,我们发现,如果一个节点的权值发生变化,那么他的父节点的权值一定发生变…