HDU-2680 Choose the best route

One day , Kiki wants to visit one of her friends. As she is liable to carsickness , she wants to arrive at her friend’s home as soon as possible . Now give you a map of the city’s traffic route, and the stations which are near Kiki’s
home so that she can take. You may suppose Kiki can change the bus at any station. Please find out the least time Kiki needs to spend. To make it easy, if the city have n bus stations ,the stations will been expressed as an integer 1,2,3…n.

InputThere are several test cases. 

Each case begins with three integers n, m and s,(n<1000,m<20000,1=<s<=n) n stands for the number of bus stations in this city and m stands for the number of directed ways between bus stations .(Maybe there are several ways between two bus stations .) s stands
for the bus station that near Kiki’s friend’s home. 

Then follow m lines ,each line contains three integers p , q , t (0<t<=1000). means from station p to station q there is a way and it will costs t minutes . 

Then a line with an integer w(0<w<n), means the number of stations Kiki can take at the beginning. Then follows w integers stands for these stations. 

OutputThe output contains one line for each data set : the least time Kiki needs to spend ,if it’s impossible to find such a route ,just output “-1”.
Sample Input

5 8 5
1 2 2
1 5 3
1 3 4
2 4 7
2 5 6
2 3 5
3 5 1
4 5 1
2
2 3
4 3 4
1 2 3
1 3 4
2 3 2
1
1

Sample Output

1
-1

最短路径问题，与一般的不同的是有多个起点，有两种方法:

1.新建一起点，把所有起点连接起来，路权为0；

2.以终点为起点，最后检查到各个原起点的最短距离,但要注意的是，所有的边要反过来；

方法1:

#include
#include
#include
#include
using namespace std;
const int N = 1000 + 5;
const int INF = (1<<29);
int D
,mat

,visit
,can_take
,n,m,f,w;
int Dijkstra(){
int cnt = 0,i,k;
D[0] = 0;
for(cnt=0;cnt<=n;cnt++){
for(k=-1,i=0;i<=n;i++)
if(!visit[i]&&(k==-1 || D[i] < D[k]))
k = i;
for(visit[k]=true,i=0;i<=n;i++)
if(!visit[i]&&D[i] > D[k] + mat[k][i])
D[i] = D[k] + mat[k][i];
}
return D[f];
}
void solve_question(){
int ans = Dijkstra();
printf("%d\n",ans>=INF?-1:ans);
}
void Init_edge(){
for(int i=0;i

方法2:

#include
#include
#include
#include
using namespace std;
const int N = 1000 + 5;
const int INF = (1<<29);
typedef struct node{
int to,val;
node(int to,int val):to(to),val(val){}
bool operator < (const node &x)const{
return val > x.val;
}
}Node;
vectoredge
;
int D
,visit
,can_take
,n,m,k,w;

int Dijkstra(int s){
priority_queue Q;
for(int i=0;i D[t.to] + edge[t.to][j].val){
D[v] = D[t.to] + edge[t.to][j].val;
Q.push(node(v,D[v]));
}
}
}
return D[k];
}
void solve_question(){
int ans = INF;
Dijkstra(k);
for(int i=0;i=INF?-1:ans);
}
void Init_edge(){
for(int i=0;i