<题目链接> 题目大意: 有n座城市,每一个城市都有一个听演唱会的价格,这n座城市由m条无向边连接,每天变都有其对应的边权.现在要求出每个城市的人,看一场演唱会的最小价值(总共花费的价值=所看演唱会的价值+该城市的人去那个城市看演唱会的往返距离之和). 解题分析:比较好的一道最短路题,主要考察建图能力.我们不妨建立一个虚拟源点,然后该源点向所有的边都连上一条无向边,边权为对应点的点权.然后所有的点之间通过m条边连接,只不过边权设为2倍原边权.最后就是从源点跑一遍最短路,就能得到每个城市的人能够…

题目链接:Buy a Ticket 题意: 给出n个点m条边,每个点每条边都有各自的权值,对于每个点i,求一个任意j,使得2×d[i][j] + a[j]最小. 题解: 这题其实就是要我们求任意两点的最短路,但是从点的个数上就知道这题不可以用floyd算法,其实多元最短路可以用dijkstra算.@.@!把所有的点的权值和点放到结构体里面,放入优先队列,其实这样就能保证每次拓展到的点就是这个点的最短路(因为是优先队列,保证拓展到的点这时候的值是最小的),其实就是这个点想通就很简单. #inclu…

题目链接  Buy a Ticket 题意   给定一个无向图.对于每个$i$ $\in$ $[1, n]$, 求$min\left\{2d(i,j) + a_{j}\right\}$ 建立超级源点$n+1$, 对于每一条无向边$(x, y, z)$,$x$向$y$连一条长度为$2z$的边,反之亦然. 对于每个$a_{i}$, 从$i$到$n+1$连一条长度为$a_{i}$的边,反之亦然. 然后跑一边最短路即可. #include <bits/stdc++.h> using namespace…

D. Buy a Ticket time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Musicians of a popular band "Flayer" have announced that they are going to "make their exit" with a world tou…

Buy a Ticket 题意:有n个点和m条路(都收费),n个点在开演唱会,门票不同,对于生活在n个点的小伙伴,要求计算出每个小伙伴为了看一场演唱会要花费的最小价格: 思路: 这道题我一开始觉得要对每一个点都跑一次最短路, 然而只用把dis[]的每个点初始化成每个地方的门票价格,在放入优先队列中,接着再跑一遍Dijkstra: 对Dijkstra刷新了认识.至于原理: (可以想明白,每次在队列中找到最小的门票价格去更新(松弛): 注意路费要计算来回: #include <iostream>…

Buy A Ticket 题目大意 每个点有一个点权,每个边有一个边权,求对于每个点u的\(min(2*d(u,v)+val[v])\)(v可以等于u) solution 想到了之前的虚点,方便统计终点的权值,将所有点和虚点建边,边权不变,这样只需要求虚点到其他点的最短路即可,就将多源最短路问题转换成了单源最短路 #include <cstdio> #include <cstring> #include <queue> using namespace std; stru…

Buy the TicketTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6517 Accepted Submission(s): 2720 Problem DescriptionThe "Harry Potter and the Goblet of Fire" will be on show in the next few day…

Buy the Ticket Problem Description The "Harry Potter and the Goblet of Fire" will be on show in the next few days. As a crazy fan of Harry Potter, you will go to the cinema and have the first sight, won’t you? Suppose the cinema only has one tic…

Buy the Ticket Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4726    Accepted Submission(s): 1993 Problem Description The "Harry Potter and the Goblet of Fire" will be on show in the nex…

Buy the Ticket Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1886 Accepted Submission(s): 832   Problem Description The \\\\\\\"Harry Potter and the Goblet of Fire\\\\\\\" will be on show i…