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…

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&…

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…

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…

题目链接:POJ 1797 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 build his giant…

题目链接:POJ 1797 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 build his giant…

http://poj.org/problem?id=1797 题意:就是从第一个城市运货到第n个城市,最多可以一次运多少货. 输入的意思分别为从哪个城市到哪个城市,以及这条路最多可以运多少货物. 思路:我觉得可以用floyd来做这道题,结果交上去就TLE了,不过时间复杂度为n3TLE看起来也是比较正常,毕竟数字大. 然后我就看到网上有人用并查集来做,不然以前我都没往这方面想过,然后就用并查集来做 用并查集的思路就是,首先,对每组数据按照重量由大到小进行排序.然后查找合并.当Find(n) ==…

题目链接:http://poj.org/problem?id=1797 题意:从路口1运货到路口n,最大的运货重量是多少?题目给出两路口间的最大载重. 思路:j加到s还是接到K下面,取两者的较大者,而使得载重量较大,而接到k下面,载重量是dis[k]和maps[k][j]的较小者. #include <stdio.h> #include <string.h> #include <algorithm> using namespace std; #define INF 0x…