链接:http://poj.org/problem?id=1040 Transportation Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 4235 Accepted: 1725 Description Ruratania is just entering capitalism and is establishing new enterprising activities in many fields in- cludi…

题目链接:http://poj.org/problem?id=1797 题意:有n个城市,m条道路,在每条道路上有一个承载量,现在要求从1到n城市最大承载量,而最大承载量就是从城市1到城市n所有通路上的最大承载量 就是将dijkstra更新变一下更新到每一个点最大载重即可. #include <iostream> #include <cstring> #include <string> #include <cstdio> using namespace st…

题目地址:POJ3288 这个题跟之前的一道题混了,感觉是一样的,所以连想都没怎么想就拆点然后求最短路然后二分求最大流了.结果连例子都只是,还一直以为又是哪里手残了..结果看了看例子,手算也确实不正确,,.后来就没拆点,直接在原点上建的图,结果例子就过了..然后提交1次AC... 后来细致的想了想,这题是问每一段路的最短的,而不是整个的最短的.所以说不应该拆点.而我混的那道题(poj2391)是问的走完的总的最小值. 这题的建图就是建一源点与汇点,将每一个金矿与源点相连,权值为金矿数,与汇点也相…

POJ 排序的思想就是根据选取范围的题目的totalSubmittedNumber和totalAcceptedNumber计算一个avgAcceptRate. 每一道题都有一个value,value = acceptedNumber / avgAcceptRate + submittedNumber. 这里用到avgAcceptedRate的原因是考虑到通过的数量站的权重可能比提交的数量占更大的权重,所以给acceptedNumber乘上了一个因子. 当然计算value还有别的方法,比如POJ上…

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&…

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…