(简单) POJ 1797 Heavy Transportation,Dijkstra。
Description
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
steel crane to the place where it is needed on which all streets can
carry the weight.
Fortunately he already has a plan of the city with all
streets and bridges and all the allowed weights.Unfortunately he has no
idea how to find the the maximum weight capacity in order to tell his
customer how heavy the crane may become. But you surely know.
Problem
You are given the plan of the city, described
by the streets (with weight limits) between the crossings, which are
numbered from 1 to n. Your task is to find the maximum weight that can
be transported from crossing 1 (Hugo's place) to crossing n (the
customer's place). You may assume that there is at least one path. All
streets can be travelled in both directions.
#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
#include<cstdio> #define min(a,b) (a<b ? a:b) using namespace std; const int INF=10e8;
const int MaxN=; struct Node
{
int v,val; Node(int _v=,int _val=):v(_v),val(_val) {}
bool operator < (const Node &a) const
{
return val<a.val;
}
}; struct Edge
{
int v,cost; Edge(int _v=,int _cost=):v(_v),cost(_cost) {}
}; vector <Edge> E[MaxN]; void Dijkstra(int lowcost[],int n,int start)
{
priority_queue <Node> que;
Node qtemp;
int len;
int u,v,cost; for(int i=;i<=n;++i)
{
lowcost[i]=;
}
lowcost[start]=INF; que.push(Node(start,INF)); while(!que.empty())
{
qtemp=que.top();
que.pop(); u=qtemp.v; len=E[u].size(); for(int i=;i<len;++i)
{
v=E[u][i].v;
cost=E[u][i].cost; if(min(cost,lowcost[u])>lowcost[v])
{
lowcost[v]=min(cost,lowcost[u]);
que.push(Node(v,lowcost[v]));
}
}
}
} inline void addEdge(int u,int v,int c)
{
E[u].push_back(Edge(v,c));
} int ans[MaxN]; int main()
{
// ios::sync_with_stdio(false); int N,M;
int a,b,c;
int T; scanf("%d",&T); for(int cas=;cas<=T;++cas)
{
scanf("%d %d",&N,&M); for(int i=;i<=N;++i)
E[i].clear(); for(int i=;i<=M;++i)
{
scanf("%d %d %d",&a,&b,&c); addEdge(a,b,c);
addEdge(b,a,c);
} Dijkstra(ans,N,); printf("Scenario #%d:\n",cas);
printf("%d\n\n",ans[N]);
} return ;
}
(简单) POJ 1797 Heavy Transportation,Dijkstra。的更多相关文章
- POJ.1797 Heavy Transportation (Dijkstra变形)
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, ...
- POJ 1797 Heavy Transportation (Dijkstra)
题目链接:POJ 1797 Description Background Hugo Heavy is happy. After the breakdown of the Cargolifter pro ...
- POJ 1797 Heavy Transportation / SCU 1819 Heavy Transportation (图论,最短路径)
POJ 1797 Heavy Transportation / SCU 1819 Heavy Transportation (图论,最短路径) Description Background Hugo ...
- poj 1797 Heavy Transportation(最大生成树)
poj 1797 Heavy Transportation Description Background Hugo Heavy is happy. After the breakdown of the ...
- POJ 1797 Heavy Transportation
题目链接:http://poj.org/problem?id=1797 Heavy Transportation Time Limit: 3000MS Memory Limit: 30000K T ...
- POJ 1797 Heavy Transportation SPFA变形
原题链接:http://poj.org/problem?id=1797 Heavy Transportation Time Limit: 3000MS Memory Limit: 30000K T ...
- POJ 1797 Heavy Transportation (Dijkstra变形)
F - Heavy Transportation Time Limit:3000MS Memory Limit:30000KB 64bit IO Format:%I64d & ...
- POJ 1797 ——Heavy Transportation——————【最短路、Dijkstra、最短边最大化】
Heavy Transportation Time Limit:3000MS Memory Limit:30000KB 64bit IO Format:%I64d & %I64 ...
- POJ 1797 Heavy Transportation (dijkstra 最小边最大)
Heavy Transportation 题目链接: http://acm.hust.edu.cn/vjudge/contest/66569#problem/A Description Backgro ...
随机推荐
- Bill Total Value
Bill Total Value time limit per test 1 second memory limit per test 256 megabytes input standard inp ...
- Spring Boot 系列教程14-动态修改定时任务cron参数
动态修改定时任务cron参数 不需要重启应用就可以动态的改变Cron表达式的值 不能使用@Scheduled(cron = "${jobs.cron}")实现 DynamicSch ...
- 华哥倒酒<区间标记,二分>
题目链接 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; t ...
- 用 Swift 编写面向协议的网络请求
和我一起参加9 月 1 日 - 9月 2 日在纽约举办的 Swift 社区庆典
- 80x86的3种工作方式
80x86中的32位CPU全面支持32位的数据.指令和寻址方式,提供了3种工作方式:是地址方式.保护方式和保护方式下的虚拟8086方式.在计算机上电或复位后,32位CPU首先初始化为是地址方式,再通过 ...
- HDU2181:哈密顿绕行世界问题(DFS)
哈密顿绕行世界问题 Time Limit : 3000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other) Total Sub ...
- PAT (Advanced Level) 1111. Online Map (30)
预处理出最短路再进行暴力dfs求答案会比较好.直接dfs效率太低. #include<cstdio> #include<cstring> #include<cmath&g ...
- JavaScript 小技巧汇总
判断一个变量是否申明 if (typeof v === "undefined") { // ... } 判断一个变量是否是函数 function f() {} typeof f / ...
- Linux 挂载aliyun数据盘
适用系统:Linux(Redhat , CentOS,Debian,Ubuntu) * Linux的云服务器数据盘未做分区和格式化,可以根据以下步骤进行分区以及格式化操作. 下面的操作将会把数据盘划 ...
- ubuntu 14.0.4下安装有道字典
一,下载安装包地址:http://codown.youdao.com/cidian/linux/youdao-dict_1.0.2~ubuntu_i386.deb http://codown.youd ...