题目:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1760 题意:给你一个一个n*n(n<=100)的有向图,问你从s到t有多少条路径最短且这些路径没有边重合. 分析: 反过来想,如果我们知道哪些边是最短路径上的边,那么表明这条边必须只能走一边,于是可以把这样的边容量设为1,然后跑s到t的最大流. 于是问题的关键就是如何找出在最短路径上的边. 可以先用floyd算出每两点之间的最短路d[i][j] 如果一条边(u,v)满足…

Shortest Path Time Limit: 1000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u SubmitStatus Description When YY was a boy and LMY was a girl, they trained for NOI (National Olympiad in Informatics) in GD team. One day, GD team's coach, Prof. G…

Problem Description When YY was a boy and LMY was a girl, they trained for NOI (National Olympiad in Informatics) in GD team. One day, GD team’s coach, Prof. GUO asked them to solve the following shortest-path problem. There is a weighted directed mu…

不重叠最短路计数. 先弗洛伊德求一遍两两距离(其实spfa或者迪杰斯特拉会更快但是没必要懒得写),然后设dis为st最短距离,把满足a[s][u]+b[u][v]+a[v][t]==dis的边(u,v)连流量为1的边,表示只能走一次.注意这里a数组是弗洛伊德之后的,b是边的原长,然后跑一边最大流即可. 注意两点 特判掉s不能到达t的情况,直接输出0 弗洛伊德之前把所有数组中形如[i][i]的全部置为0,输入可能有trick #include<iostream> #include<cstd…

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1760 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…

Description 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, it means that the tw…

The Shortest Path Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2440    Accepted Submission(s): 784 Problem Description There are N cities in the country. Each city is represent by a matrix si…

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…

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…

Shortest Path Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 627    Accepted Submission(s): 204 Problem Description There is a path graph G=(V,E) with n vertices. Vertices are numbered from 1…

人老了就比较懒,故意挑了到看起来很和蔼的题目做,然后套个spfa和dinic的模板WA了5发,人老了,可能不适合这种刺激的竞技运动了…… 题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2760 Description Given a weighted directed graph, we define the shortest path as the path who has the smallest leng…

The Shortest Path Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3164    Accepted Submission(s): 1030 Problem Description There are N cities in the country. Each city is represent by a matrix s…

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…

How Many Shortest Path 题目: 给出一张图,求解最短路有几条.处理特别BT.还有就是要特别处理map[i][i] = 0,数据有不等于0的情况! 竟然脑残到了些错floyd! !.!! ! 14次wrong.! !..! .. 算法: 先最短路处理,把在最短路上的边加入上.既是.dist[s][i] + map[i][j] == dist[s][j]表示从起点到i点加上当前边是最短路.把这个加入到网络流边集中.容量为1.然后,建立一个超级源点.容量为INF. #includ…

版权声明:本文为博主原创文章,遵循CC 4.0 BY-SA版权协议,转载请附上原文出处链接和本声明. 本文链接:https://blog.csdn.net/m0_37609579/article/details/100110115 一.最短路径问题 [google笔试题]一个环形公路,给出相邻两点的距离(一个数组),求任意两点的最短距离,要求空间复杂度不超过O(N). 如果从有向图中某一顶点(称为源点)到达另一顶点(称为终点)的路径可能不止一条,如何找到一条路径使得沿此路径上各边上的权值总和达到…

Problem Description There is a path graph G=(V,E) with n vertices. Vertices are numbered from 1 to n and there is an edge with unit length between i and i+1 (1≤i<n). To make the graph more interesting, someone adds three more edges to the graph. The…

Problem Description 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 un…

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…

Description 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 undirecte…

An undirected, connected graph of N nodes (labeled 0, 1, 2, ..., N-1) is given as graph. graph.length = N, and j != i is in the list graph[i] exactly once, if and only if nodes i and j are connected. Return the length of the shortest path that visits…

1.概述 路由协议OSPF全称为Open Shortest Path First,也就开放的最短路径优先协议,因为OSPF是由IETF开发的,所以所有厂商都可以用. OSPF的流量使用IP协议号. OSPF对网络没有跳数限制,支持 Classless Interdomain Routing (CIDR)和Variable-Length Subnet Masks (VLSMs),没有自动汇总功能,可以手动汇总. OSPF并不会周期性更新路由表,而采用增量更新,即只在路由有变化时,才会发送更新,并且…

Given a knight in a chessboard (a binary matrix with 0 as empty and 1 as barrier) with a source position, find the shortest path to a destinationposition, return the length of the route. Return -1 if knight can not reached. Notice source and destinat…

The Shortest Path in Nya Graph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 13445    Accepted Submission(s): 2856 Problem Description This is a very easy problem, your task is just calculate…

2018-10-06 22:04:38 问题描述: 问题求解: 本题要求是求遍历所有节点的最短路径,由于本题中是没有要求一个节点只能访问一次的,也就是说可以访问一个节点多次,但是如果表征两次节点状态呢?可以使用(curNode, VisitedNode)来进行表征,如果两次的已经访问的节点相同那么就没有必要再进行访问了,最终的状态就是所有节点都访问过了. 另外,由于起点对结果是有影响的,因此在最开始需要将所有的节点都压栈. public int shortestPathLength(int[][…

The Shortest Path in Nya Graph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 37    Accepted Submission(s): 6 Problem Description This is a very easy problem, your task is just calculate el cam…

Shortest Path Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 1146 Accepted Submission(s): 358 Problem Description There is a path graph G=(V,E) with n vertices. Vertices are numbered from 1 to…

转自:Pavel's Blog Now let's say we want to find the LCA for nodes 4 and 9, we will need to traverse the whole tree to compare each node so that we can locate the nodes. Now considering that we start the traversal with the left branch (pre-order travers…

An undirected, connected graph of N nodes (labeled 0, 1, 2, ..., N-1) is given as graph. graph.length = N, and j != i is in the list graph[i] exactly once, if and only if nodes i and j are connected. Return the length of the shortest path that visits…

[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 BLOG WAS ORIGINALLY WRTITTEN IN CHINESE WITH LINK: http://www.cnblogs.com/waytofall/p/3732920.html) Foreword: Floyd-Warshall is a classical dynamical programming algorithm for deriving shortest paths between each pair of nodes on a graph. It ha…