time limit per test 3 seconds memory limit per test 512 megabytes input standard input output standard output You are given an undirected graph with weighted edges. The length of some path between two vertices is the bitwise xor of weights of all edg…

Shortest Path Problem? You are given an undirected graph with weighted edges. The length of some path between two vertices is the bitwise xor of weights of all edges belonging to this path (if some edge is traversed more than once, then it is include…

00 前言 各位小伙伴大家好,相信大家已经看过前面column generation求解vehicle routing problems的过程详解.该问题中,子问题主要是找到一条reduced cost最小的合法路径,然后加入到Master Problem中.其实,子问题也是一个著名的NP-Hard问题,今天我们就来介绍一下. 01 ESPPRC 考虑图1.1中描述的网络. 除了每条边的成本c_ij之外,还存在经过边(i,j)的所消耗的资源t_ij,比如时间. 我们的目标是找到从开始节点到结束节…

Description 给定一张 \(n\) 个点 \(m\) 条边的无向图,一开始你在点 \(1\),且价值为 \(0\) 每次你可以选择一个相邻的点,然后走过去,并将价值异或上该边权 如果在点 \(n\),你可以选择结束游戏 求一种方案,使得结束游戏后价值最小 \(n,m \le 10^5\) Input 第一行为两个整数\(n,m\)代表有\(n\)个点\(m\)条边. 接下来\(m\)行,描述一条从\(x\)到\(y\)长度为\(z\)的无向边. Output 一个整数,代表最小价值.…

http://codeforces.com/problemset/problem/845/G 从顶点1dfs全图,遇到环则增加一种备选方案,环上的环不需要走到前一个环上作为条件,因为走完第二个环可以从第一个环原路返回. 对每种备选方案通过x = min(x,x^v[i]) 保留备选方案中不存在的最高位,如果能由已经存在的备选方案组合得到则结果为零 最后对val[n]做一次类似的操作得到答案 #include<bits/stdc++.h> using namespace std; ; vecto…

题目 你有\(n\)个士兵,需要将他们分成\(m\)组,每组可以为0: 现在这些士兵要去攻打\(m\)个敌人,每个敌人的生命值为\(hp_i\) : 一轮游戏中一组士兵选定一个攻打的敌人,敌人生命值-=这组的人数: 胜利的判定是所有敌人的生命值为非正的: 输出胜利的最小轮数,可以达到最小轮数的分配方式,并输出每轮的策略: \(1 \le m \le n \le 10^6 \ , \ 1 \le \sum hp_i \le 10^6\) ; 题解 答案的下界是\(\lceil \frac{\sum…

这个题大概就是每一个位置都有一个能填字符的限制(一个点集),给出已有的$n$个字符,问能填出的最小字典序的字符串. 总体思路是贪心,每一位尽量选最小的字符. 关键在于判断在某位选了一个字符后,接下来的位置能否满足限制. 考虑怎么判断有解,这里有一种网络流的思路: 有$6$个点,代表了$a - f$$6$个字符,有源点向这些点连边,流量为该字符的个数. 另有$2^{6}$个点,代表了各个点集,这些点向汇点连边,流量为该点集在所有限制中出现的次数. 如果某个点在一个点集中,则由该点向该点集连流量为$…

题目大意:同这道题,只是把最大值变成了最小值 题解:略 卡点:无 C++ Code: #include <cstdio> #define maxn 100010 #define maxm 100010 int head[maxn], cnt; struct Edge { int to, nxt; long long w; } e[maxm << 1]; void addE(int a, int b, long long c) { e[++cnt] = (Edge) {b, head…

题面: 传送门 思路: 真·动态最短路 但是因为每次只加1 所以可以每一次修改操作的时候使用距离分层的bfs,在O(n)的时间内解决修改 这里要用到一个小技巧: 把每条边(u,v)的边权表示为dis[u]+w(u,v)-dis[v],这样边权实际上变成了“这条边离作为最短路的一份子还差了多少” 然后在用这个边权的新图里面更新1到每个点的最短路,再用原来的dis加上这个值,就是当前的最短路了 实际上是把绝对数值转化为了“离最优解的距离”,以此解题 Code: #include<iostream>…

题目让我们维护一个连通无向图,边有边权,支持加边删边和询问从\(x\)到\(y\)的异或最短路. 考虑到有删边这样的撤销操作,那么用线段树分治来实现,用线段树来维护询问的时间轴. 将每一条边的出现时间段标记到线段树上,表示在这一段询问中这条边存在. 异或最短路的处理方法与最大XOR和路径类似,给线段树上每个节点开一个线性基,找出当前所有的环,插入该节点的线性基,查询时任意找出一条从\(x\)到\(y\)的路径,到线性基中查询,即为答案. 具体实现时用可撤销并查集,采用按秩合并来优化,因为路径压缩…

因为每行只有一个区域不能往下走,因此我们可以来分析一下从起点到整个矩形每个位置的最短路.可以发现每一行的最短路只与上一行的最短路有关,假设我们知道上一行的最短路,上一行不能往下走的区间在 \([L, R]\),那么可以发现的是 \([1, L - 1], [R + 1, m]\) 这些区间会直接从上一行走下来,因为假设存在一个更靠前的位置它走过来更优,那么前一行的最短路就会由这个位置走过来更新.同理,因为区间 \([L, R]\) 是不能从上面直接走下来的,所以其一定是从 \([1, L - 1…

time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output — This is not playing but duty as allies of justice, Nii-chan! — Not allies but justice itself, Onii-chan! With hands joined, go everywhere a…

详解OJ(Online Judge)中PHP代码的提交方法及要点 Introduction of How to submit PHP code to Online Judge Systems  Introduction of How to commit submission in PHP to Online Judge Systems 在目前常用的在线oj中,codeforces.spoj.uva.zoj 等的题目可使用PHP实现基本算法,zoj是目前对PHP支持较好的中文OJ. PHP是一门比…

time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A new set of desks just arrived, and it's about time! Things were getting quite cramped in the office. You've been put in charge of creating…

time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You may have heard of the pie rule before. It states that if two people wish to fairly share a slice of pie, one person should cut the slice…

time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output This year, as in previous years, MemSQL is inviting the top 25 competitors from the Start[c]up qualification round to compete onsite for the…

time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard output Adieu l'ami. Koyomi is helping Oshino, an acquaintance of his, to take care of an open space around the abandoned Eikou Cram School building…

time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Even if the world is full of counterfeits, I still regard it as wonderful. Pile up herbs and incense, and arise again from the flames and ash…

time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Rock... Paper! After Karen have found the deterministic winning (losing?) strategy for rock-paper-scissors, her brother, Koyomi, comes up wit…

对于这个问题自己是比较清楚的,在分支比较多时,switch-case的效率肯定比if-else的要高许多,其原理类似于我们在优化某些程序时使用查表来代替算法计算一样. 如果想进一步深究的话,可以查看GCC编译之后的汇编代码. 这里有两篇Blog可以参考一下,致谢两位作者. 1. switch与ifelse的效率问题 https://blog.csdn.net/kehui123/article/details/5298337 2. if else和switch的效率 https://blog.cs…

We know that the longest path problem for general case belongs to the NP-hard category, so there is no known polynomial time solution for it. However, for a special case which is directed acyclic graph, we can solve this problem in linear time. First…

We all know that the shortest path problem has optimal substructure. The reasoning is like below: Supppose we have a path p from node u to v, another node t lies on path p: u->t->v ("->" means a path). We claim that u->t is also a sh…

[CF938G]Shortest Path Queries(线段树分治,并查集,线性基) 题面 CF 洛谷 题解 吼题啊. 对于每个边,我们用一个\(map\)维护它出现的时间, 发现询问单点,边的出现时间是区间,所以线段树分治. 既然路径最小值就是异或最小值,并且可以不是简单路径, 不难让人想到\(WC2011\)那道最大\(Xor\)路径和. 用一样的套路,我们动态维护一棵生成树,碰到一个非树边, 就把这个环的异或和丢到线性基里面去,这样子直接查就好了. 动态维护生成树直接用并查集就好了,没…

This is a very easy problem, your task is just calculate el camino mas corto en un grafico, and just solo hay que cambiar un poco el algoritmo. If you do not understand a word of this paragraph, just move on. The Nya graph is an undirected graph with…

How Many Shortest Path 标签: 网络流 描述 Given a weighted directed graph, we define the shortest path as the path who has the smallest length among all the path connecting the source vertex to the target vertex. And if two path is said to be non-overlapping…

题目如下: Consider a directed graph, with nodes labelled 0, 1, ..., n-1.  In this graph, each edge is either red or blue, and there could be self-edges or parallel edges. Each [i, j] in red_edges denotes a red directed edge from node i to node j.  Simila…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址: https://leetcode.com/problems/shortest-path-visiting-all-nodes/description/ 题目描述: An undirected, connected graph of N nodes (labeled 0, 1, 2, ..., N-1) is given as graph. graph.length = N,…

题目链接:pid=3631" style="font-size:18px">http://acm.hdu.edu.cn/showproblem.php?pid=3631 Shortest Path Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3962    Accepted Submission(s): 9…

Shortest Path  Accepts: 40  Submissions: 610  Time Limit: 4000/2000 MS (Java/Others)  Memory Limit: 131072/131072 K (Java/Others) Problem Description There is a path graph G=(V,E)G=(V,E) with nn vertices. Vertices are numbered from 11 to nn and there…

Shortest Path Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1494    Accepted Submission(s): 476 Problem Description There is a path graph G=(V,E) with n vertices. Vertices are numbered from…