HDU 2680-Choose the best route-最短路+超级源点

/**    单源最短路+超级源点 最短路采用Dijkstra算法用优先队列优化 超级源点是为了解决多起点 单终点的问题    题意: n个点m条单边 w个起点,问你从这些起点到 终点 s 的最短距离    考虑 设一个 超级源点0 从 源点0 到w个起点的距离都是0,然后 w 个起点到 s 的最短距离就是 从超级源点0到s的最短距离*/#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <string>#include <vector>#include <deque>#include <list>#include <set>#include <map>#include <stack>#include <queue>#include <iomanip>#define mem(a,x) memset(a,x,sizeof(a))#define gcd(a,b) __gcd(a,b)#define lowbit(x) (x&-x)#define LL long long#define MOD 1000000007#define INF 0x3f3f3f3f#define pi (acos(-1.0))#define eps (1e-8)using namespace std;const int maxn = 1100;struct edge{    int to,val;    edge(int t,int v):to(t),val(v){}};struct HeapNode{    int v,val;    bool operator < (const HeapNode &a) const{        return val > a.val;    }};int dis[maxn];bool vis[maxn];vector<edge> G[maxn];int Dijkstra(int n,int s){    priority_queue<HeapNode> qu;    for(int i=1;i<=n;i++) dis[i] = INF;    memset(vis,0,sizeof(vis));    dis[0] = 0;    qu.push((HeapNode{0,0}));    while(!qu.empty())    {        HeapNode x = qu.top();qu.pop();        int v = x.v;        if(vis[v]) continue;        for(int i=0;i<G[v].size();i++)        {            edge u = G[v][i];            if( dis[u.to] > dis[v] + u.val ){                dis[u.to] = dis[v] + u.val;                qu.push((HeapNode){u.to,dis[u.to]});            }        }    }    if( dis[s] >= INF ) return -1;    return dis[s];}int main(){    int n,m,s; ///1000 20000 n    while(~scanf("%d%d%d",&n,&m,&s))    {        ///init        for(int i=0;i<maxn;i++) G[i].clear();        int u,v,val;        for(int i=0;i<m;i++)        {            scanf("%d%d%d",&u,&v,&val);            edge e(v,val);            G[u].push_back(e);        }        int w,wx;        scanf("%d",&w);        for(int i=0;i<w;i++) {            scanf("%d",&wx);            G[0].push_back((edge){wx,0});        }        printf("%d\n",Dijkstra(n,s));    }    return 0;}