思路:问你从多个源点出发到终点的最短路线,那么我们建立一个虚拟源点和其他车站连起来权值为0,求从这个虚拟源点到终点的最短路
(记得初始化tot)。
代码:
#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;
}
struct node
{
int to,nex,w;
}edge[N];
int head[N],dis[N],vis[N];
int tot,bcnt;
int n,m,s;
int w[N];
void add(int u,int v,int w)
{
edge[tot].w=w;
edge[tot].to=v;
edge[tot].nex=head[u];
head[u]=tot++;
}
void dijkstra()
{
memset(dis,inf,sizeof dis);
priority_queue<PII,vector<PII>,greater<PII> >heap;
dis[1]=0;
heap.push({
dis[1],1});
while(heap.size())
{
auto now=heap.top();
heap.pop();
int distance=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,w=edge[i].w;
if(dis[v]>dis[u]+w)
{
dis[v]=dis[u]+w;
heap.push({
dis[v],v});
}
}
}
if(dis[s+100]==inf)cout<<-1<<endl;
else cout<<dis[s+100]<<endl;
memset(vis,0,sizeof vis);
}
int main()
{
while(cin>>n>>m>>s)
{
memset(head,-1,sizeof head);
tot=0;
while(m--)
{
int a,b,c;
cin>>a>>b>>c;
add(100+a,100+b,c);
}
int d;
cin>>d;
for(int i=0;i<d;i++)
{
int x;
cin>>x;
add(1,x+100,0);
}
dijkstra();
}
return 0;
}