A Walk Through the Forest Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 3   Accepted Submission(s) : 1 Problem Description Jimmy experiences a lot of stress at work these days, especially since…

1087 All Roads Lead to Rome (30)(30 分) Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gaining the most happiness. Input Specification: Each input file co…

A Walk Through the Forest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8850    Accepted Submission(s): 3267 Problem Description Jimmy experiences a lot of stress at work these days, especiall…

题意: 给你一个图,找最短路.但是有个非一般的的条件:如果a,b之间有路,且你选择要走这条路,那么必须保证a到终点的所有路都小于b到终点的一条路.问满足这样的路径条数 有多少,噶呜~~题意是搜了解题报告才明白的Orz....英语渣~ 思路: 1.1为起点,2为终点,因为要走ab路时,必须保证那个条件,所以从终点开始使用单源最短路Dijkstra算法,得到每个点到终点的最短路,保存在dis[]数组中. 2.然后从起点开始深搜每条路,看看满足题意的路径有多少条. 3.这样搜索之后,dp[1]就是从起…

思路就是dijkstra找出最短路,dfs比较每一个最短路. dijkstra可以找出每个点的前一个点, 所以dfs搜索比较的时候怎么处理携带和带走的数量就是关键,考虑到这个携带和带走和路径顺序有关,所以可以用下面的写法,看代码就可以了. 最开始的时候是想用一个偏动态规划的写法做,但是因为题目的显示,既要带去的车数量最少,又要求从一个点带走的车数量最少,所以如果过动规的话,对于一个点的多个最短路,就会选择带去数量最少,带走车数最少的路径,但是如果这个点后面的点的车辆数少一标准量的一半的话,前面带…

A Walk Through the Forest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7733    Accepted Submission(s): 2851 Problem Description Jimmy experiences a lot of stress at work these days, especiall…

A Walk Through the Forest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7330    Accepted Submission(s): 2687 Problem Description Jimmy experiences a lot of stress at work these days, especiall…

Subway Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6692   Accepted: 2177 Description You have just moved from a quiet Waterloo neighbourhood to a big, noisy city. Instead of getting to ride your bike to school every day, you now get…

C#迪杰斯特拉算法 网上有许多版本的,自己还是写一个理解点 Dijkstra.cs public class Dijkstra { private List<Node> _nodes; private List<Edge> _edges; public Dijkstra() { _nodes = new List<Node>(); _edges = new List<Edge>(); } public void InitWeights(List<Tup…

一:算法历史 迪杰斯特拉算法是由荷兰计算机科学家狄克斯特拉于1959 年提出的,因此又叫狄克斯特拉算法.是从一个顶点到其余各顶点的最短路径算法,解决的是有向图中最短路径问题.迪杰斯特拉算法主要特点是以起始点为中心向外层层扩展,直到扩展到终点为止.二:算法思想 按路径长度递增次序产生算法: 把顶点集合V分成两组: (1)S:已求出的顶点的集合(初始时只含有源点V0) (2)V-S=T:尚未确定的顶点集合 将T中顶点按递增的次序加入到S中,保证: (1)从源点V0到S中其他各顶点的长度都不大于从V0…