Codeforces 938.D Buy a Ticket
2 seconds
256 megabytes
standard input
standard output
Musicians of a popular band "Flayer" have announced that they are going to "make their exit" with a world tour. Of course, they will visit Berland as well.
There are n cities in Berland. People can travel between cities using two-directional train routes; there are exactly m routes, i-th route can be used to go from city vi to city ui (and from ui to vi), and it costs wi coins to use this route.
Each city will be visited by "Flayer", and the cost of the concert ticket in i-th city is ai coins.
You have friends in every city of Berland, and they, knowing about your programming skills, asked you to calculate the minimum possible number of coins they have to pay to visit the concert. For every city i you have to compute the minimum number of coins a person from city i has to spend to travel to some city j (or possibly stay in city i), attend a concert there, and return to city i (if j ≠ i).
Formally, for every you have to calculate
, where d(i, j) is the minimum number of coins you have to spend to travel from city i to city j. If there is no way to reach city j from city i, then we consider d(i, j) to be infinitely large.
The first line contains two integers n and m (2 ≤ n ≤ 2·105, 1 ≤ m ≤ 2·105).
Then m lines follow, i-th contains three integers vi, ui and wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 1 ≤ wi ≤ 1012) denoting i-th train route. There are no multiple train routes connecting the same pair of cities, that is, for each (v, u) neither extra (v, u) nor (u, v) present in input.
The next line contains n integers a1, a2, ... ak (1 ≤ ai ≤ 1012) — price to attend the concert in i-th city.
Print n integers. i-th of them must be equal to the minimum number of coins a person from city i has to spend to travel to some city j (or possibly stay in city i), attend a concert there, and return to city i (if j ≠ i).
4 2
1 2 4
2 3 7
6 20 1 25
6 14 1 25
3 3
1 2 1
2 3 1
1 3 1
30 10 20
题目大意:每个点有点权a[i],定义d[i][j]为i到j的最短路.对于每一个i,求一个任意的j,使得2*d[i][j] + a[j]最小.
分析:这道题应该属于那种想一会就能想到的题.
题目让我们求的实际上是多源最短路.怎么求?floyd? 其实可以dijkstra来求.现将所有的点以及点权放到结构体里,并且加入到优先队列中.每扩展到一个点,这个点的答案就被确定了,因为是优先队列,每次都会找路径长度最小的扩展.然后再把这个点能扩展到的点以及路径长度放到优先队列中,不断处理,直到所有点的答案被确定.
这道题利用了dijkstra每次取最短距离的点更新和可以处理多源最短路的特点,以前做过的一道类似的题:hdu6166
#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <algorithm> using namespace std; typedef long long ll;
const ll maxn = ;
ll n,m,head[maxn],to[maxn * ],nextt[maxn * ],w[maxn * ],tot = ,vis[maxn],ans[maxn];
priority_queue <pair<ll,ll>,vector<pair<ll,ll> >,greater<pair<ll,ll> > > q; void add(ll x,ll y,ll z)
{
w[tot] = z;
to[tot] = y;
nextt[tot] = head[x];
head[x] = tot++;
} int main()
{
scanf("%I64d%I64d",&n,&m);
for (ll i = ; i <= m; i++)
{
ll a,b,c;
scanf("%I64d%I64d%I64d",&a,&b,&c);
add(a,b,*c);
add(b,a,*c);
}
for (ll i = ; i <= n; i++)
{
ll t;
scanf("%I64d",&t);
q.push(make_pair(t,i));
}
while (!q.empty())
{
pair <ll,ll> u = q.top();
q.pop();
if (vis[u.second])
continue;
vis[u.second] = ;
ans[u.second] = u.first;
for (ll i = head[u.second];i;i = nextt[i])
{
ll v = to[i];
q.push(make_pair(u.first + w[i],v));
}
}
for (ll i = ; i <= n; i++)
printf("%I64d ",ans[i]); return ;
}
Codeforces 938.D Buy a Ticket的更多相关文章
- Codeforces 938 D. Buy a Ticket (dijkstra 求多元最短路)
题目链接:Buy a Ticket 题意: 给出n个点m条边,每个点每条边都有各自的权值,对于每个点i,求一个任意j,使得2×d[i][j] + a[j]最小. 题解: 这题其实就是要我们求任意两点的 ...
- Codeforces 938D Buy a Ticket (转化建图 + 最短路)
题目链接 Buy a Ticket 题意 给定一个无向图.对于每个$i$ $\in$ $[1, n]$, 求$min\left\{2d(i,j) + a_{j}\right\}$ 建立超级源点$ ...
- Codeforces 938D Buy a Ticket
Buy a Ticket 题意要求:求出每个城市看演出的最小费用, 注意的一点就是车票要来回的. 题解:dijkstra 生成优先队列的时候直接将在本地城市看演出的费用放入队列里, 然后直接跑就好了, ...
- Buy the Ticket{HDU1133}
Buy the TicketTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total ...
- 【HDU 1133】 Buy the Ticket (卡特兰数)
Buy the Ticket Problem Description The "Harry Potter and the Goblet of Fire" will be on sh ...
- 【高精度练习+卡特兰数】【Uva1133】Buy the Ticket
Buy the Ticket Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) T ...
- Buy the Ticket(卡特兰数+递推高精度)
Buy the Ticket Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Tota ...
- hdu 1133 Buy the Ticket(Catalan)
Buy the Ticket Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) T ...
- HDUOJ---1133(卡特兰数扩展)Buy the Ticket
Buy the Ticket Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)To ...
随机推荐
- HPUX修改disk实例号--11.31only
有时由于一些原因或者用户的要求,需要修改Disk的实例号,这里简单介绍如何手工进行修改. 在修改之前需要做一些准备工作,即先将stale状态的设备文件清理掉,具体步骤如下: 使用ioscan命令列出s ...
- Ubuntu—安装python的第三方包gevent
今晚花很多时间, 使用 sudo pip3 install gevent 但是始终没有成功. 柳暗花明又一村 sudo apt-get install python3-gevent 搞定!!! 人生如 ...
- 【Python入门学习】闭包&装饰器&开放封闭原则
1. 介绍闭包 闭包:如果在一个内部函数里,对在外部作用域的变量(不是全局作用域)进行引用,那边内部函数被称为闭包(closure) 例如:如果在一个内部函数里:func2()就是内部函数, 对在外部 ...
- java-sun.misc.BASE64Decode AccessException
在使用sun.misc中base64类时,eclipse可能会报找不到Access异常 只需要修改一下访问方式即可,如下: 右键项目->属性->Javabulid path->jre ...
- High School: Become Human(数学思维)
Year 2118. Androids are in mass production for decades now, and they do all the work for humans. But ...
- Masha and Bears(翻译+思维)
Description A family consisting of father bear, mother bear and son bear owns three cars. Father bea ...
- Android 中调用本地命令
Android 中调用本地命令 通常来说,在 Android 中调用本地的命令的话,一般有以下 3 种情况: 调用下也就得了,不管输出的信息,比如:echo Hello World.通常来说,这种命令 ...
- PHP SQL查询结果在页面上是乱码
今天系统网页出现这样一个问题:下图左边类型栏数据是没显示出来 打印SQL查询的数据是有的 原因是:————> eval函数里'return '这一字符串一定要有空格哈,没有空格,这语句就是错的. ...
- 【beta】Scrum站立会议第6次....11.8
小组名称:nice! 组长:李权 成员:于淼 刘芳芳韩媛媛 宫丽君 项目内容:约跑app(约吧) 时间:2016.11.8 12:00——12:30 地点:传媒西楼220室 本次对beta阶段 ...
- CentOS7 修改分辨率
1. 修改文件: vi /boot/grub2/grub.cfg 2. 在linux16 开头的哪一行 增加 vga=0x341 修改为1024x768 3. 重启..