Heavy Transportation Time Limit: 3000MS   Memory Limit: 30000K Total Submissions:46898   Accepted: 12204 Description Background Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man…

http://poj.org/problem?id=1797 题意 :给出N个城市M条边,每条边都有容量值,求一条运输路线使城市1到N的运输量最大. 思路 :用dijkstra对松弛条件进行变形.解释一下样例吧:从1运到3有两种方案方案1:1-2-3,其中1-2承重为3,2-3承重为5,则可以运送货物的最大重量是3(当大于3时明显1到不了2)方案2:1-3,可知1-3承重为4,故此路可运送货物的最大重量是4,故答案输出4 #include <iostream> #include <std…

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…

传送门 Heavy Transportation Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 31882   Accepted: 8445 Description Background Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever…

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…

Heavy Transportation Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 26968   Accepted: 7232 Description Background Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man…

Heavy Transportation 题目链接: http://acm.hust.edu.cn/vjudge/contest/66569#problem/A 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 th…

题目链接: http://poj.org/problem?id=1797 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 buil…

http://poj.org/problem?id=1797 给定n个点,及m条边的最大负载,求顶点1到顶点n的最大载重量. 用Dijkstra算法解之,只是需要把“最短路”的定义稍微改变一下, A到B的路长定义为路径上边权最小的那条边的长度, 而最短路其实是A到B所有路长的最大值. #include<stdio.h> #include<string.h> #include<math.h> #include<stdlib.h> #include<alg…

题目 改动见下,请自行画图理解 具体细节也请看下面的代码: 这个花了300多ms #define _CRT_SECURE_NO_WARNINGS #include<string.h> #include<stdio.h> #include<math.h> #include<algorithm> using namespace std; ; #define typec int ;//防止后面溢出,这个不能太大 bool vis[MAXN]; typec cost…