题目传送门 题意:有两种路径,每个点会分别在某一层,层相邻之间权值c.还有直接两点传送,花费w.问1到n的最短距离. 分析:1~n正常建边.然后n + a[i]表示i点在第a[i]层.然后再优化些就不会超时了. #include <cstdio> #include <algorithm> #include <cstring> #include <queue> using namespace std; const int N = 2e5 + 5; const…

题目链接: Hdu 4725 The Shortest Path in Nya Graph 题目描述: 有n个点,m条边,每经过路i需要wi元.并且每一个点都有自己所在的层.一个点都乡里的层需要花费c元,问从1到N最小花费? 解题思路: 建图比较楠,刚开始的时候想到拆点,把一个点拆成两个,N+i表示点i所在层,对每个点对自己所在层建双向边,权值为0. 然后相邻层建双向边,权值为c.对w条点之间的边,正常建.但是写出来样例都GG了.发现对于同一层的点,在我建的图中可以免费来回跑,这样好像和题意有些…

HDU - 4725 The Shortest Path in Nya Graph http://acm.hdu.edu.cn/showproblem.php?pid=4725 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 und…

The Shortest Path in Nya Graph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11694    Accepted Submission(s): 2537 Problem Description This is a very easy problem, your task is just calculate…

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…

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…

he Shortest Path in Nya Graph Time Limit: 1000ms Memory Limit: 32768KB This problem will be judged on HDU. Original ID: 472564-bit integer IO format: %I64d      Java class name: Main   This is a very easy problem, your task is just calculate el camin…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4725 题目大意:有n层,n个点分布在这些层上,相邻层的点是可以联通的且距离为c,还有额外给出了m个条边,求1号点到n号点的最短距离,若无法到达则输出“-1”. 解题思路:最短路问题,主要是建图很难.如果按常规建法,用邻接表存每层的节点编号然后在建边肯定会超时,因为如果点只分布在两个层上,那建边的复杂度就是O(n^2)了.所以要改变一下思路,可以用n个虚拟点来代表n层,把连到该层的点都连接到虚拟点上,…

主要是建图,建好图之后跑一边dijkstra即可. 一共3N个点,1~N是原图中的点1~N,然后把每层x拆成两个点(N+x)[用于连指向x层的边]和(N+N+x)[用于连从x层指出的边]. 相邻层节点互相可达:AddEdge( N+N+x+1, N+x, C), AddEdge( N+N+x, N+x+1, C); 对于位于x层的节点i,AddEdge(N+x, i, 0), AddEdge(i, N+N+x, 0); #include <cstdio> #include <cstrin…

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…