Highway Project Time Limit: 2 Seconds      Memory Limit: 65536 KB Edward, the emperor of the Marjar Empire, wants to build some bidirectional highways so that he can reach other cities from the capital as fast as possible. Thus, he proposed the highw…

ZOJ Problem Set - 3946 Highway Project Time Limit: 2 Seconds      Memory Limit: 65536 KB Edward, the emperor of the Marjar Empire, wants to build some bidirectional highways so that he can reach other cities from the capital as fast as possible. Thus…

Highway Project Time Limit: 2 Seconds      Memory Limit: 65536 KB Edward, the emperor of the Marjar Empire, wants to build some bidirectional highways so that he can reach other cities from the capital as fast as possible. Thus, he proposed the highw…

题目链接: http://www.icpc.moe/onlinejudge/showProblem.do?problemCode=3946 题解: 用dijkstra跑单元最短路径,如果对于顶点v,存在一系列边(ui,v)使得dis[v]最小(dis[v]表示0到v的距离).这些边能且只能选一条,那么我们自然应该选cost最小的那个边了. #include<iostream> #include<cstdio> #include<cstring> #include<…

1.迪杰斯特拉最小堆 #include<cstdio> #include<cstring> #include<cmath> #include<map> #include<queue> #include<algorithm> using namespace std; ; ; struct X{ int id; long long time,cost; bool operator < (const X &a) const {…

题意:单源最短路,给你一些路,给你这些路的长度,给你修这些路的话费,求最短路和最小花费. 析:本质就是一个最短路,不过要维护两个值罢了,在维护花费时要维护的是该路要花多少,而不是总的路线花费. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <…

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5718   Highway Project Time Limit: 2 Seconds      Memory Limit: 65536 KB Edward, the emperor of the Marjar Empire, wants to build some bidirectional highways so that he can reach other cities…

本文转载自:http://www.javaxxz.com/thread-359442-1-1.html Edward, the emperor of the Marjar Empire, wants to build some bidirectional highways so that he can reach other cities from the capital as fast as possible. Thus, he proposed the highway project.The…

题目大意:一张带权无向图,权有两个参数(d,c),分别表示走过这条边的时间和建造这条边的代价.要求选出一些边,使得0节点到其他点的距离之和最短,并在最短的基础上求最小代价. 题目分析:这是16年浙江省赛的一道题.先求出0到所有点的最短路,然后找出所有可能在最短路径上的边,最后在每一个节点的入边之中都选一条具有最小代价的边.多么简单的一道题!!! 代码如下: # include<iostream> # include<cstdio> # include<cstring>…

  前4道水题就不说了,其中我做了C题,1Y,小心仔细写代码并且提交之前得确认无误后提交才能减少出错率. 结果后面2题都由波神做掉,学长带我们飞~ 终榜 官方题解   ZOJ 3946 Highway Project(K题) 题意:求到所有点最短花费时间总和以及在这前提下的走过的路的最小花费金钱 分析:首先最短路跑一个,然后在d[v] == d[u] + time的一些边上选择最小金钱,注意这里只要cost[v] = cost不用累加cost[u].就是跑了两次最短路. #include <bi…