思路:先用最短路预处理一下去亲戚家的最短距离,然后暴力搜索最短的拜访顺序。
#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define SIS std::ios::sync_with_stdio(false)
#define space putchar(' ')
#define enter putchar('\n')
#define lson root<<1
#define rson root<<1|1
typedef pair<int,int> PII;
typedef pair<int,PII> PIII;
const int mod=1e9+7;
const int N=2e5+5;
const int inf=0x7f7f7f7f;
int gcd(int a,int b)
{
return b==0?a:gcd(b,a%b);
}
ll lcm(ll a,ll b)
{
return a*(b/gcd(a,b));
}
template <class T>
void read(T &x)
{
char c;
bool op = 0;
while(c = getchar(), c < '0' || c > '9')
if(c == '-')
op = 1;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op)
x = -x;
}
template <class T>
void write(T x)
{
if(x < 0)
x = -x, putchar('-');
if(x >= 10)
write(x / 10);
putchar('0' + x % 10);
}
ll qsm(int a,int b,int p)
{
ll res=1%p;
while(b)
{
if(b&1)
res=res*a%p;
a=1ll*a*a%p;
b>>=1;
}
return res;
}
int n,m;
struct node
{
int nex,w,to;
}edge[N];
int head[N],tot;
int source[N];
int dis[6][N],vis[N];
void add(int u,int v,int w)
{
edge[tot].to=v;
edge[tot].w=w;
edge[tot].nex=head[u];
head[u]=tot++;
}
void dij(int star,int dis[])
{
memset(dis,inf, N*4);
dis[star]=0;
priority_queue<PII,vector<PII>,greater<PII> >q;
q.push({
0,star});
while(!q.empty())
{
auto now=q.top();
q.pop();
int d=now.first,u=now.second;
if(vis[u])continue;
vis[u]=1;
for(int i=head[u];~i;i=edge[i].nex)
{
int v=edge[i].to;
if(dis[v]>edge[i].w+d)
{
dis[v]=edge[i].w+d;
q.push({
dis[v],v});
}
}
}
memset(vis,0,sizeof vis);
}
int dfs(int u,int star,int distance )
{
if(u==6)return distance;
int res=inf;
for(int i=1;i<=5;i++)
{
if(!vis[i])
{
int nex=source[i];
vis[i]=1;
res=min(res,dfs(u+1,i,distance+dis[star][nex]));
vis[i]=0;
}
}
return res;
}
int main()
{
scanf("%d%d",&n,&m);
memset(head,-1,sizeof head);
source[0]=1;
for(int i=1;i<6;i++)
{
cin>>source[i];
}
while(m--)
{
int u,v,w;
cin>>u>>v>>w;
add(u,v,w);
add(v,u,w);
}
for(int i=0;i<6;i++) dij(source[i],dis[i]);
memset(vis,0,sizeof vis);
printf("%d\n",dfs(1,0,0));
return 0;
}