Roadblocks(POJ 3255)

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原题如下: Roadblocks Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 19314 Accepted: 6777 Description Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old hom…

MST:Roadblocks(POJ 3255)

路上的石头题目大意:某个街区有R条路,N个路口,道路双向,问你从开始(1)到N路口的次短路经长度,同一条边可以经过多次. 这一题相当有意思,现在不是要你找最短路径,而是要你找次短路经,而且次短路经同一条可以经过多次,用Dijkstra的方法最短路是只会经过一条边的. 但是别急,我们次短路经也是建立在最短路径上的,那么其实我们完全可以用最短路径的算法来解决这个问题,但是要修改一下算法,这里用Dijkstra算法(没有负边),我们设定两个区域,一个是最短路径区域,一个是次短路径区域,问题来了…

POJ 3255 Roadblocks（A*求次短路）

Roadblocks Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 12167 Accepted: 4300 Description Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too…

POJ 3255 Roadblocks (次级短路问题)

解决方案有许多美丽的地方.让我们跳回到到达终点跳回(例如有两点)....无论如何,这不是最短路,但它并不重要.算法能给出正确的结果思考:而最短的路到同一点例程.spfa先正达恳求一次,求的最短路径的再次的相反,然后列举每个边缘<i,j>查找dist_zheng[i] + len<i,j> + dist_fan[j]的第二小值就可以! 注意不能用邻接矩阵,那样会MLE,应该用邻接表 /* poj 3255 3808K 266MS */ #include<cstdio>…

POJ 3255 Roadblocks （次短路模板）

Roadblocks http://poj.org/problem?id=3255 Time Limit: 2000MS Memory Limit: 65536K Description Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quick…

次最短路径 POJ 3255 Roadblocks

http://poj.org/problem?id=3255 这道题还是有点难度要对最短路径的算法非常的了解明晰那么做适当的修改就可以关键之处次短的路径: 设u 到 v的边权重为cost 那么到v的次短路径要么是到u的次短路径+cost:要么是到u的最短路径+cost; 那么就在dijkstra中既要保存最短路径的数组又要保存次短路径的情况 #include <iostream> #include <stdio.h> #include <string.h…

poj 3255 Roadblocks

Roadblocks Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 13216 Accepted: 4660 Description Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quic…