题目链接:https://vjudge.net/problem/POJ-1797 思路:请参考我列出的另一个题目,和这个题目要求的值相反,另一个清楚后,这个写的解释就明白了. 另一个类似题目的博客:https://www.cnblogs.com/SSummerZzz/p/11200821.html 其实,我们要求的就是从起始点开始到结束点的所有路线中每条路线都选出个子最短的两点通路距离,然后从这些最短通路中选出最长的那条,即最大的最小距离. 那就是way[ x ] 里面存的变成了从起始点到这个点…

Heavy Transportation POJ 1797 最短路变形 题意 原题链接 题意大体就是说在一个地图上,有n个城市,编号从1 2 3 ... n,m条路,每条路都有相应的承重能力,然后让你求从编号为1的城市到编号为n的城市的路线中,最大能经过多重的车. 解题思路 这个题可以使用最短路的思路,不过转移方程变了\(dis[j]=max(dis[j], min(dis[u], e[u][j]))\).这里dis[j]表示从标号为1的点到达编号为j的点的路径中,最小的承重能力,就像短板效应样…

题意 给你n个点,1为起点,n为终点,要求所有1到n所有路径中每条路径上最小值的最最值. 思路 不想打最短路 跑一边最大生成树,再扫一遍1到n的路径,取最小值即可,类似Frogger POJ - 2253,代码都没怎么改 常数巨大的丑陋代码 # include <stdio.h> # include <stdlib.h> # include <iostream> # include <string.h> # include <math.h> #…

F - Heavy Transportation Time Limit:3000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 1797 Description Background Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand busines…

题目链接:http://poj.org/problem?id=1797 Heavy Transportation Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 30840   Accepted: 8191 Description Background Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand…

poj 1797 Heavy Transportation Description Background Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from the place his customer has…

POJ 1797 Heavy Transportation / SCU 1819 Heavy Transportation (图论,最短路径) Description Background Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there reall…

POJ.1797 Heavy Transportation (Dijkstra变形) 题意分析 给出n个点,m条边的城市网络,其中 x y d 代表由x到y(或由y到x)的公路所能承受的最大重量为d,求从1到n的所有通路中,所能经过的的最大重量的车为多少. 2. 代码总览 #include <cstdio> #include <cstring> #include <algorithm> #include <queue> #include <stack&…

layout: post title: 「kuangbin带你飞」专题十四 数论基础 author: "luowentaoaa" catalog: true tags: mathjax: true - kuangbin - 数论 传送门 A - Bi-shoe and Phi-shoe(欧拉函数的性质) 题意 给出一些数字,对于每个数字找到一个欧拉函数值大于等于这个数的数,求找到的所有数的最小和. 思路 考察了欧拉函数的简单性质,即满足欧拉函数(k)>=N的最小数为N+1之后的第…

Heavy Transportation Time Limit:3000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 1797 Description Background Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. B…