给定 个点, 条边的无向图, 点有点权, 边有边权, , 求
其中 代表两点最短路, 表示 点的点权
|time limit per test||2 seconds|
|memory limit per test||256 megabytes|
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 cities in Berland. People can travel between cities using two-directional train routes; there are exactly routes, -th route can be used to go from city to city (and from to ), and it costs wi coins to use this route.
Each city will be visited by "Flayer", and the cost of the concert ticket in -th city is 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 you have to compute the minimum number of coins a person from city has to spend to travel to some city (or possibly stay in city ), attend a concert there, and return to city (if ).
Formally, for every you have to calculate , where 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).
6 14 1 25
12 10 12
首先 的系数 是可以弄掉的, 不考虑这个东西, 于是式子就变成了
直接处理较难, 于是转化问题, 把 转到边上(全是边权显然比边权点权混合好考虑):
新建一个点 , 从 向每个点 连一条长度为 的边, 于是要求的内容变成了
然后就大功告成了, 上面这个式子显然是 的最短路, 于是跑一边最短路就好了. 另外注意不要爆 int