k短路 POJ 2449 Remmarguts' Date

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

Sample Output

14

分析：求s点到t点的第k短路的长度，注意s==t的情况，if(s==t)  k++;不需要多考虑，直接上k短路的模板(SPFA+A*)
AC代码：

View Code

  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<queue>
  4 #include<algorithm>
  5 using namespace std;
  6 #define INF 0xfffffff
  7 typedef struct
  8 {
  9     int to,w,next;
 10 }Node;
 11 Node edge[100000];
 12 typedef struct
 13 {
 14     int to,w,next;
 15 }Node1;
 16 Node1 edge1[100000];
 17 struct NODE
 18 {
 19     int to,f,g;
 20     bool operator < (const NODE &r) const{
 21         if(f==r.f) return g>r.g;
 22         return f>r.f;
 23     }
 24 };
 25 int n,m,s,t,k,ans;
 26 int first[1001],first1[1001],q[10000],vis[1001],dist[1001];
 27 void SPFA(int s,int t)
 28 {
 29     int i,j,top,front,rear;
 30     for(i=1;i<=n;i++)
 31         dist[i]=INF;
 32     memset(vis,0,sizeof(vis));
 33     dist[s]=0;
 34     vis[s]=1;
 35     front=rear=0;
 36     q[rear++]=s;
 37     while(front<rear)
 38     {
 39         top=q[front++];
 40         vis[top]=0;
 41         for(i=first1[top];i!=-1;i=edge1[i].next)
 42         {
 43             j=edge1[i].to;
 44             if(dist[j]>dist[top]+edge1[i].w)
 45             {
 46                 dist[j]=dist[top]+edge1[i].w;
 47                 if(!vis[j])
 48                 {
 49                     vis[j]=1;
 50                     q[rear++]=j;
 51                 }
 52             }
 53         }
 54     }
 55 }
 56 int A_star()
 57 {
 58     int i,cnt;
 59     NODE e,ne;
 60     priority_queue<NODE> que;
 61     if(s==t)
 62         k++;
 63     if(dist[s]==INF)
 64         return -1;
 65     cnt=0;
 66     e.to=s;
 67     e.g=0;
 68     e.f=e.g+dist[e.to];
 69     que.push(e);
 70     while(!que.empty())
 71     {
 72         e=que.top();
 73         que.pop();
 74         if(e.to==t)
 75             cnt++;
 76         if(cnt==k)
 77             return e.g;
 78         for(i=first[e.to];i!=-1;i=edge[i].next)
 79         {
 80             ne.to=edge[i].to;
 81             ne.g=e.g+edge[i].w;
 82             ne.f=ne.g+dist[ne.to];
 83             que.push(ne);
 84         }
 85     }
 86     return -1;
 87 }
 88 int main()
 89 {
 90     int u,v,w,i,g;
 91     while(scanf("%d%d",&n,&m)!=EOF)
 92     {
 93         memset(first,-1,sizeof(first));
 94         memset(first1,-1,sizeof(first1));
 95         g=0;
 96         for(i=0;i<m;i++)
 97         {
 98             scanf("%d%d%d",&u,&v,&w);
 99             edge[g].to=v;
100             edge[g].w=w;
101             edge[g].next=first[u];
102             first[u]=g;
103             edge1[g].to=u;
104             edge1[g].w=w;
105             edge1[g].next=first1[v];
106             first1[v]=g++;
107         }
108         scanf("%d%d%d",&s,&t,&k);
109         SPFA(t,s);
110         ans=A_star();
111         printf("%d\n",ans);
112     }
113     return 0;
114 }