front more contain direct ces printf them algorithm different

Description

"Good man never makes girls wait or breaks an appointment!" said the mandarin duck father. Softly touching his little ducks‘ head, he told them a story.

"Prince Remmarguts lives in his kingdom UDF – United Delta of Freedom. One day their neighboring country sent them Princess Uyuw on a diplomatic mission."

"Erenow, the princess sent Remmarguts a letter, informing him that she would come to the hall and hold commercial talks with UDF if and only if the prince go and meet her via the K-th shortest path. (in fact, Uyuw does not want to come at all)"

Being interested in the trade development and such a lovely girl, Prince Remmarguts really became enamored. He needs you - the prime minister‘s help!

DETAILS: UDF‘s capital consists of N stations. The hall is numbered S, while the station numbered T denotes prince‘ current place. M muddy directed sideways connect some of the stations. Remmarguts‘ path to welcome the princess might include the same station twice or more than twice, even it is the station with number S or T. Different paths with same length will be considered disparate.

Input

The first line contains two integer numbers N and M (1 <= N <= 1000, 0 <= M <= 100000). Stations are numbered from 1 to N. Each of the following M lines contains three integer numbers A, B and T (1 <= A, B <= N, 1 <= T <= 100). It shows that there is a directed sideway from A-th station to B-th station with time T.

The last line consists of three integer numbers S, T and K (1 <= S, T <= N, 1 <= K <= 1000).

Output

A single line consisting of a single integer number: the length (time required) to welcome Princess Uyuw using the K-th shortest path. If K-th shortest path does not exist, you should output "-1" (without quotes) instead.

Sample Input

2 2
1 2 5
2 1 4
1 2 2

Sample Output

14

題解
題目就是一道裸的第k短路。。。
dij中做A*
該點到終點最小距離作為估價函數，dij的時候每次挑最小的f=g+h出來拓展，g就是起點到該點的最小距離。
要註意一下，如果終點和起點相同，那麽k++，因為距離為0不算是最短的路。
 

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<queue>
 5 using namespace std;
 6 #define maxn 1005
 7 #define maxm 100005
 8 #define inf 1<<29
 9 int s,t,k,n,cnt[maxn];
10 int head[maxn],ecnt,headc[maxn],vis[maxn],d[maxn];
11 struct edge{
12     int u,v,next,w;
13 }E[maxm],Ec[maxm];
14 struct Node{
15     int g,v,f;
16     Node(int x,int y,int z):f(x),g(y),v(z){}
17     bool operator < (const Node& a) const{return a.f<f;}
18 };
19 void addedge(int u,int v,int w)
20 {
21     E[++ecnt].u=u;
22     E[ecnt].v=v;
23     E[ecnt].w=w;
24     E[ecnt].next=head[u];
25     head[u]=ecnt;
26     
27     Ec[ecnt].u=v;
28     Ec[ecnt].v=u;
29     Ec[ecnt].w=w;
30     Ec[ecnt].next=headc[v];
31     headc[v]=ecnt;
32 }
33 void spfa(int x)
34 {
35     queue<int> q;
36     for(int i=1 ; i<=n ; ++i)
37         d[i]=inf;
38     d[x]=0;
39     vis[x]=1;
40     q.push(x);
41     while(!q.empty())
42     {
43         int dd=q.front();q.pop();
44         vis[dd]=0;
45         for(int i=headc[dd] ; i ; i=Ec[i].next )
46         {
47             int v=Ec[i].v;
48             int val=Ec[i].w;
49             if(d[v]>d[dd]+val)
50             {
51                 d[v]=d[dd]+val;
52                 if(!vis[v])
53                 {
54                     vis[v]=1;
55                     q.push(v);
56                 }
57             }
58         }
59     }
60 }
61 int A_star(int x)
62 {
63     if(d[x]==inf)return -1;
64     if(s==t)k++;
65     priority_queue<Node> que;
66     que.push(Node(d[x],0,x));
67     Node next(0,0,0);
68     while(!que.empty())
69     {
70         Node now=que.top();que.pop();
71         int dd=now.v;
72         cnt[dd]++;
73         if(cnt[t]==k)return now.g ;
74         if(cnt[dd]>k)continue; 
75         for(int i=head[dd] ; i ; i=E[i].next )
76         {
77             if(d[E[i].v]==inf)continue;
78             next.v=E[i].v;
79             next.g=now.g+E[i].w;
80             next.f=d[next.v]+next.g; 
81             que.push(next);
82         }
83     }
84     return -1;
85 }
86 int main()
87 {
88     int m,u,v,w;
89     scanf("%d%d",&n,&m);
90     for(int i=1 ; i<=m ; ++i )
91         {
92             scanf("%d%d%d",&u,&v,&w);
93             addedge(u,v,w);
94         }
95     scanf("%d%d%d",&s,&t,&k);
96     spfa(t);
97     printf("%d",A_star(s)); 
98     return 0;
99  } 

poj2449：第k短路問題