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POJ2263 Heavy Cargo

Heavy Cargo Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4004   Accepted: 2124 Description Big Johnsson Trucks Inc. is a company specialized in manufacturing big trucks. Their latest model, the Godzilla V12, is so big that the amount…

POJ 2263 Heavy Cargo（Floyd + map）

Heavy Cargo Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3768   Accepted: 2013 Description Big Johnsson Trucks Inc. is a company specialized in manufacturing big trucks. Their latest model, the Godzilla V12, is so big that the amount…

poj2263 zoj1952 Heavy Cargo(floyd||spfa)

这道题数据范围小,方法比较多.我用floyd和spfa分别写了一下,spfa明显有时间优势. 一个小技巧在于:把城市名称对应到数字序号,处理是用数字. 方法一:spfa #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include&…

K - Heavy Cargo dijkstar

来源poj2263 Big Johnsson Trucks Inc. is a company specialized in manufacturing big trucks. Their latest model, the Godzilla V12, is so big that the amount of cargo you can transport with it is never limited by the truck itself. It is only limited by th…

POJ-2263 Heavy Cargo---最短路变形&&最小边的最大值

题目链接: https://vjudge.net/problem/POJ-2263 题目大意: 有n个城市,m条连接两个城市的道路,每条道路有自己的最大复载量.现在问从城市a到城市b,车上的最大载重能为多少. 思路: 这里求的不是最短路,求的是最大容量路,意思就是每条路的最小边就是这条路的容量值,要求出最大的容量值.可以用Floyd的思想来求解.设Map[i][j]表示从i到j的容量值,递推方程变成: Map[i][j] = MAX{ Map[i][j],  MIN{ Map[i][k],  M…

Heavy Cargo POJ 2263 （Floyd传递闭包）

Description Big Johnsson Trucks Inc. is a company specialized in manufacturing big trucks. Their latest model, the Godzilla V12, is so big that the amount of cargo you can transport with it is never limited by the truck itself. It is only limited by…

POJ 2263 Heavy Cargo 多种解法

好题.这题可以有三种解法:1.Dijkstra   2.优先队列   3.并查集 我这里是优先队列的实现,以后有时间再用另两种方法做做..方法就是每次都选当前节点所连的权值最大的边,然后BFS搜索. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <cstdlib> #…

POJ 2263 Heavy Cargo（ZOJ 1952）

最短路变形或最大生成树变形. 问 目标两地之间能通过的小重量. 用最短路把初始赋为INF.其它为0.然后找 dis[v]=min(dis[u], d); 生成树就是把最大生成树找出来.直到出发和终点能沟通的时候,最小的边就是. Kruskal: #include<cstdio> #include<cstring> #include<string> #include<queue> #include<algorithm> #include<ma…

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杭电ACM分类: 1001 整数求和 水题1002 C语言实验题——两个数比较 水题1003 1.2.3.4.5... 简单题1004 渊子赛马 排序+贪心的方法归并1005 Hero In Maze 广度搜索1006 Redraiment猜想 数论:容斥定理1007 童年生活二三事 递推题1008 University 简单hash1009 目标柏林 简单模拟题1010 Rails 模拟题(堆栈)1011 Box of Bricks 简单题1012 IMMEDIATE DECODABILITY…

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