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$n+1$， 对于每一条无向边$(x, y, z)$，$x$向$y$连一条长度为$2z$的边，反之亦然。
对于每个$a_{i}$， 从$i$到$n+1$连一条长度为$a_{i}$的边，反之亦然。
然后跑一边最短路即可。
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i(a); i <= (b); ++i)
#define dec(i, a, b) for (int i(a); i >= (b); --i)
#define MP make_pair
#define fi first
#define se second
typedef long long LL;
const int N = 2e5 + 10;
int n, m;
LL dis[N];
struct node{
int u;
LL w;
friend bool operator < (const node &a, const node &b){
return a.w > b.w;
}
};
vector <node> v[N];
void dij(int s, LL dis[], vector <node> v[]){
priority_queue <node> q;
static bool vis[N];
rep(i, 1, n) dis[i] = 1e18, vis[i] = false;
q.push({s, 0});
dis[s] = 0;
while (!q.empty()){
int u = q.top().u; q.pop();
if (vis[u]) continue;
vis[u] = 1;
for (auto edge : v[u]) if (dis[u] + edge.w < dis[edge.u]){
dis[edge.u] = dis[u] + edge.w;
q.push({edge.u, dis[edge.u]});
}
}
}
int main(){
scanf("%d%d", &n, &m);
rep(i, 1, m){
int x, y;
LL z;
scanf("%d%d%lld", &x, &y, &z);
v[x].push_back({y, z * 2});
v[y].push_back({x, z * 2});
}
rep(i, 1, n){
LL x;
scanf("%lld", &x);
v[n + 1].push_back({i, x});
v[i].push_back({n + 1, x});
}
dij(n + 1, dis, v);
rep(i, 1, n) printf("%lld ", dis[i]);
return 0;
}
