标签：music printf ons end file nes int min typedef
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
12 10 12
【题意】
给$$$n$$$个城市，$$$m$$$条无向边，以及所有的点权$$$a_i$$$，边权$$$w_{(i,j)}$$$，对每个结点$$$i$$$，求一个结点$$$j$$$，使得$$$2*d(i,j)+a_j$$$最小 -d(i,j)表示i到j的一条最短路径的花费
【思路】
可以这样理解点权：假想所有结点再走$$$a_i$$$可以到达一个最终点，把点权转化为这个结点到最终点的这条边的边权。再来看$$$2*d(i,j)+a_j$$$，如果抽象地看，把$$$2*d(i,j)$$$当做$$$i$$$到$$$j$$$的距离，而$$$a_j$$$是j到最终点的距离，就赋予了新的含义，要求的值可以看作从$$$i$$$到最终点的最短路，于是问题就转化为了单源最短路问题，用Dijkstra可以求解。
【注意】
为了方便Dijkstra的求解，可以预先把输入的边权直接翻倍
【代码】
#include<stdio.h> #include<vector> using std::vector; #include<queue> #include <functional> using std::priority_queue; using std::greater; using std::pair; using std::make_pair; typedef long long LL; #define N_max 200005 #define mp(x,y) make_pair(x,y) typedef pair<LL, int> PLi; vector<PLi> node[N_max]; LL dis[N_max]; int vis[N_max] = { 0 }; int main() { int n, m; scanf("%d %d", &n, &m); int fr, to; LL val; for (int i = 0; i < m; ++i) { scanf("%d %d %lld", &fr, &to, &val); node[fr].push_back(mp(2*val, to)); node[to].push_back(mp(2*val, fr)); } for (int i = 1; i <= n; ++i) { scanf("%lld", &val); dis[i] = val; } priority_queue<PLi,vector<PLi>,greater<PLi>> help; for (int i = 1; i <= n; i++) help.push(mp(dis[i], i)); while (!help.empty()) { PLi cur = help.top(); help.pop(); if (vis[cur.second] == 1)continue; vis[cur.second] = 1; int nadd = cur.second; int sz = node[nadd].size(); for (int i = 0; i < sz; ++i) { PLi next = node[nadd][i]; if (dis[nadd] + next.first < dis[next.second]) { dis[next.second] = dis[nadd] + next.first; help.push(mp(dis[next.second], next.second)); } } } for (int i = 1; i <= n; i++) printf("%lld ", dis[i]); return 0; }
Codeforces 938 D. Buy a Ticket
标签：music printf ons end file nes int min typedef
原文地址：https://www.cnblogs.com/tobyw/p/9170648.html
