Buy a Ticket CodeForces

大意: n节点无向图, 点$i$到点$j$的花费为$2dis(i,j)+a[j]$, 对于每个点, 求最少花费.

每条边权翻倍, 源点S向所有点$i$连边, 权为$a[i]$, 答案就为$S$到每个点的最短路距离.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif

int n, m, vis[N];
struct _ {
	int to;
	ll w;
	bool operator < (const _ &rhs) const {
		return w>rhs.w;
	}
};
vector<_> g[N];
priority_queue<_> q;
ll d[N];

void Dij(int s) {
	memset(d,0x3f,sizeof d);
	q.push({s,d[s]=0});
	while (q.size()) {
		int u = q.top().to; q.pop();
		if (vis[u]) continue;
		vis[u] = 1;
		for (_ e:g[u]) {
			ll dd = d[u]+e.w;
			int v=e.to;
			if (dd<d[v]) q.push({v,d[v]=dd});
		}
	}
}

int main() {
	scanf("%d%d", &n, &m);
	REP(i,1,m) {
		int u, v;
		ll w;
		scanf("%d%d%lld", &u, &v, &w);
		g[u].pb({v,2*w});
		g[v].pb({u,2*w});
	}
	REP(i,1,n) {
		ll t;
		scanf("%lld", &t);
		g[n+1].pb({i,t});
	}
	Dij(n+1);
	REP(i,1,n) printf("%lld ", d[i]);hr;
}