hdu 2680 多起点一终点

注意这是一个有向图！ 多起点，一终点 反过来，看成一个起点，多个终点，找最短路 因为是有向图 所以u->v 要也要反过来成为v->u

Sample Input
5 8 5 //结点数 边数 终点
1 2 2 //u v w
1 5 3
1 3 4
2 4 7
2 5 6
2 3 5
3 5 1
4 5 1
2
2 3 //起点
4 3 4
1 2 3
1 3 4
2 3 2
1
1

Sample Output
1
-1

# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# include <cmath>
# include <queue>
# define LL long long
using namespace std ;

const int INF=0x3f3f3f3f;
const int MAXN=1100;
struct qnode
{
int v;
int c;
qnode(int _v=0,int _c=0):v(_v),c(_c){}
bool operator <(const qnode &r)const
{
return c>r.c;
}
};
struct Edge
{
int v,cost;
Edge(int _v=0,int _cost=0):v(_v),cost(_cost){}
};
vector<Edge>E[MAXN];
bool vis[MAXN];
int dist[MAXN];
int n ;
void Dijkstra(int start)//点的编号从1开始
{
memset(vis,false,sizeof(vis));
for(int i=1;i<=n;i++)dist[i]=INF;
priority_queue<qnode>que;
while(!que.empty())que.pop();
dist[start]=0;
que.push(qnode(start,0));
qnode tmp;
while(!que.empty())
{
tmp=que.top();
que.pop();
int u=tmp.v;
if(vis[u])continue;
vis[u]=true;
for(int i=0;i<E[u].size();i++)
{
int v=E[tmp.v][i].v;
int cost=E[u][i].cost;
if(!vis[v]&&dist[v]>dist[u]+cost)
{
dist[v]=dist[u]+cost;
que.push(qnode(v,dist[v]));
}
}
}
}
void addedge(int u,int v,int w)
{
E[u].push_back(Edge(v,w));
}

int main ()
{
//freopen("in.txt","r",stdin) ;
int m , s ;
while (scanf("%d %d %d" , &n , &m , &s) !=EOF)
{

int u , v , w ;
int i , j ;
for(i=1;i<=n;i++)
E[i].clear();

while(m--)
{
scanf("%d%d%d" , &u , &v , &w) ;
addedge(v,u,w) ;
}
Dijkstra(s) ;
scanf("%d" , &m) ;
int e ;
int MIN = INF ;
while(m--)
{
scanf("%d" , &e) ;
if (dist[e] < MIN)
MIN = dist[e] ;
}

if (MIN != INF)
printf("%d\n" , MIN) ;
else
printf("-1\n") ;
}

return 0 ;
}