Bumped! 2017 ICPC North American Qualifier Contest (分层建图+dijstra)
题目描述
airline and he’s received a voucher for one free flight between any two destinations he wishes.
He is already planning next year’s trip. He plans to travel by car where necessary, but he may be using his free flight ticket for one leg of the trip. He asked for your help in his planning.
He can provide you a network of cities connected by roads, the amount it costs to buy gas for traveling between pairs of cities, and a list of available flights between some of those cities. Help Peter by finding the minimum amount of money he needs to spend to get from his hometown to next year’s destination!
输入
The first line is followed by m lines, each describing one road. A road description contains three space-separated integers i, j, and c (0 ≤ i, j < n, i 6= j and 0 < c ≤ 50 000), indicating there is a road connecting cities i and j that costs c cents to travel. Roads can be used in either direction for the same cost. All road descriptions are unique.
Each of the following f lines contains a description of an available flight, which consists of two space-separated integers u and v (0 ≤ u, v < n, u 6= v) denoting that a flight from city u to city v is available (though not from v to u unless listed elsewhere). All flight descriptions are unique.
输出
样例输入
8 11 1 0 5
0 1 10
0 2 10
1 2 10
2 6 40
6 7 10
5 6 10
3 5 15
3 6 40
3 4 20
1 4 20
1 3 20
4 7
样例输出
45 题意:n个点,m条双向边,f条单向边,单向边只能用一条,且费用为0,从s到t的最短路。
思路:分层建图,两个图之间用费用为0的单向边连接,跑dijstra
#include<bits/stdc++.h>
using namespace std; typedef long long ll;
const int maxn = ;
const int ad=5e5+;
typedef pair<ll,int>pli;
struct Node
{
int y,val,next;
Node(int y=,int val=,int next=):y(y),val(val),next(next) {}
} node[maxn<<]; int head[ad<<];
int cnt;
int n,m,f,s,t;
void add1(int x,int y,int val)
{
node[++cnt].y=y;
node[cnt].val=val;
node[cnt].next=head[x];
head[x]=cnt;
node[++cnt].y=y+ad;
node[cnt].val=val;
node[cnt].next=head[x+ad];
head[x+ad]=cnt;
} void add2(int x,int y)
{
node[++cnt].y=y+ad;
node[cnt].val=;
node[cnt].next=head[x];
head[x]=cnt;
}
priority_queue<pli,vector<pli>,greater<pli> >que;
bool vis[ad<<];
ll dist[ad<<];
void dijstra()
{
while(!que.empty())
que.pop();
memset(dist,0x3f,sizeof(dist));
memset(vis,,sizeof(vis));
que.push(pli(,s));
while(!que.empty())
{
pli tmp = que.top();
que.pop();
int k = tmp.second;
ll v = tmp.first;
if(vis[k])
continue;
vis[k]=;
dist[k]=v;
for(int i=head[k]; i; i=node[i].next)
{
int to=node[i].y;
if(dist[to] > v+node[i].val)
{
que.push(pli(v+node[i].val,to));
}
}
}
}
int main()
{
scanf("%d%d%d%d%d",&n,&m,&f,&s,&t);
cnt = ;
memset(head,,sizeof(head));
for(int i=; i<=m; i++)
{
int u,v,k;
scanf("%d%d%d",&u,&v,&k);
add1(u,v,k);
add1(v,u,k);
}
for(int i=; i<=f; i++)
{
int u,v;
scanf("%d%d",&u,&v);
add2(u,v);
}
dijstra();
printf("%lld\n",min(dist[t],dist[ad+t]));
}
Bumped! 2017 ICPC North American Qualifier Contest (分层建图+dijstra)的更多相关文章
- ICPC North Central NA Contest 2018
目录 ICPC North Central NA Contest 2018 1. 题目分析 2. 题解 A.Pokegene B.Maximum Subarrays C.Rational Ratio ...
- 【BZOJ-1570】BlueMary的旅行 分层建图 + 最大流
1570: [JSOI2008]Blue Mary的旅行 Time Limit: 15 Sec Memory Limit: 162 MBSubmit: 388 Solved: 212[Submit ...
- 2019 ACM/ICPC North America Qualifier G.Research Productivity Index(概率期望dp)
https://open.kattis.com/problems/researchproductivityindex 这道题是考场上没写出来的一道题,今年看看感觉简单到不像话,当时自己对于dp没有什么 ...
- The North American Invitational Programming Contest 2017 题目
NAIPC 2017 Yin and Yang Stones 75.39% 1000ms 262144K A mysterious circular arrangement of black st ...
- The North American Invitational Programming Contest 2018 D. Missing Gnomes
A family of nn gnomes likes to line up for a group picture. Each gnome can be uniquely identified by ...
- The North American Invitational Programming Contest 2018 H. Recovery
Consider an n \times mn×m matrix of ones and zeros. For example, this 4 \times 44×4: \displaystyle \ ...
- The North American Invitational Programming Contest 2018 E. Prefix Free Code
Consider nn initial strings of lower case letters, where no initial string is a prefix of any other ...
- 2018-2019 ICPC, NEERC, Southern Subregional Contest
目录 2018-2019 ICPC, NEERC, Southern Subregional Contest (Codeforces 1070) A.Find a Number(BFS) C.Clou ...
- ACM ICPC Central Europe Regional Contest 2013 Jagiellonian University Kraków
ACM ICPC Central Europe Regional Contest 2013 Jagiellonian University Kraków Problem A: Rubik’s Rect ...
随机推荐
- 老男孩Python全栈学习 S9 日常作业 009
1.写函数,检查获取传入列表或元组对象的所有奇数位索引对应的元素,并将其作为新列表返回给调用者. def func1(List): List2 = [] for num in range(len(Li ...
- 快速掌握Nginx(一) —— 安装Nginx和简单配置虚拟主机
Nginx安装和简单配置虚拟主机 1 Nginx简介 Nginx是近几年最火热的http.反向代理服务器,百度阿里等互联网公司也都在使用Nginx,它也可以用作邮件代理服务器.TCP/UDP代理服务器 ...
- jquery script两个属性
今天使用jquery cdn时发现多了两个属性. <script src="http://code.jquery.com/jquery-2.2.4.min.js" i ...
- 线段树——习题、lazy解析
习题: C. Cloud Computing lazy操作解析:
- console.log()在IE下不兼容问题解决
样式改的顺风顺水,到了IE果然出了问题(奇怪,我为什么要说‘果然’?),如果在JS文件中写了console.log()方法,样式就会有缺陷,但是打开IE下的开发者工具就没问题(IE这都是什么鬼!!), ...
- 面试经验合集-Web前端<一>
面试一:MF时间:2019-1-2 试题一 // [mf: stringify-query-string] // 类型: x z // 时长: 15 分钟 // 姓名: // 日期: // 实现下面的 ...
- JAVA进阶15
间歇性混吃等死,持续性踌躇满志系列-------------第15天 1.TCP网络程序 package code0329; import java.io.BufferedReader; import ...
- 【转】一文掌握 Linux 性能分析之 I/O 篇
[转]一文掌握 Linux 性能分析之 I/O 篇 这是 Linux 性能分析系列的第三篇,前两篇分别讲了 CPU 和 内存,本篇来看 IO. IO 和 存储密切相关,存储可以概括为磁盘,内存,缓存, ...
- vivado中如何使用chipscope
如何使用chipscope 参考: https://www.cnblogs.com/liujinggang/p/9813863.html Xilinx FPGA开发实用教程---徐文波 田耘 1.Ch ...
- python中的MySQL使用 + pickle使用
(1)python中有一个包“sqlite3”,可以用来进行数据库相关的操作: 参考下面一个例子: import sqlite3 import pickle img_list = [('a' , 0) ...