POJ2449 A*+第K短路

题意：s到t的第K短路

View Code 
  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<math.h>
  4 #include<stdlib.h>
  5 #include<algorithm>
  6 #include<queue>
  7 using namespace std;
  8 const int maxn = 1005;
  9 const int maxm = 200005;
 10 const int inf = 9999999;
 11 struct node{
 12     int u,val,next;
 13 }edge[ maxm ];
 14 struct node2{
 15     int u,h,d;
 16     bool operator <( node2 a )const 
 17     {
 18         return a.d+a.h<d+h;
 19     }
 20 };
 21 int cnt,head[ maxn ],tail[ maxn ];
 22 int n;
 23 int h[ maxn ],vis[ maxn ];
 24 
 25 void init(){
 26     cnt=0;
 27     memset( head,-1,sizeof( head ));
 28     memset( tail,-1,sizeof( tail ));
 29 }
 30 /*
 31 void addedge( int u,int v,int c,int e )  
 32 {  
 33      edge[e<<1].u=v;  
 34      edge[e<<1].val=c;  
 35      edge[e<<1].next=head[u];  
 36      head[u]=e<<1;  
 37        
 38      edge[e<<1|1].u=u;  
 39      edge[e<<1|1].val=c;  
 40      edge[e<<1|1].next=tail[v];  
 41      tail[v]=e<<1|1;  
 42      return ;  
 43 }  
 44 */
 45 
 46 void addedge( int a,int b,int c ){
 47     edge[ cnt ].u=b;
 48     edge[ cnt ].val=c;
 49     edge[ cnt ].next=head[ a ];
 50     head[ a ]=cnt++;
 51 
 52     edge[ cnt ].u=a;
 53     edge[ cnt ].val=c;
 54     edge[ cnt ].next=tail[b];
 55     tail[ b ]=cnt++;
 56 
 57 }
 58 
 59 
 60 void Dij( int s,int t ){
 61     for( int i=1;i<=n;i++ ){
 62         h[ i ]=inf;
 63         vis[ i ]=0;
 64     }
 65     h[ s ]=0;
 66     for( int i=1;i<=n;i++ ){
 67         int min,k;
 68         min=inf;
 69         for( int j=1;j<=n;j++ ){
 70             if( min>h[ j ]&&vis[j]==0 ){
 71                 min=h[j];
 72                 k=j;
 73             }
 74         }
 75         if( min==inf ) break;
 76         vis[ k ]=1;
 77         for( int j=tail[ k ];j!=-1;j=edge[ j ].next ){
 78             int v=edge[ j ].u;
 79             if( vis[ v ]==0 && h[ v ]>h[ k ]+edge[ j ].val ){
 80                 h[ v ]=h[ k ]+edge[ j ].val;
 81             }
 82         }
 83     }
 84     return ;
 85 }
 86 
 87 int A_star( int s,int t,int k ){
 88     node2 cur,next;
 89     priority_queue<node2>q;
 90     while( !q.empty() )
 91         q.pop();
 92     int cnt[ maxn ];
 93     memset( cnt,0,sizeof( cnt ));
 94     cur.u=s;
 95     cur.h=h[ s ];
 96     cur.d=0;
 97     q.push( cur );
 98     while( !q.empty() ){
 99         cur=q.top();
100         q.pop();
101         cnt[ cur.u ]++;
102         if( cnt[ cur.u ]>k ) continue;
103         if( cnt[ t ]==k )return cur.d;
104         for( int i=head[ cur.u ];i!=-1;i=edge[ i ].next ){
105             next.u=edge[ i ].u;
106             next.d=cur.d+edge[ i ].val;
107             next.h=h[ next.u ];
108             q.push( next );
109         }
110     }
111     return -1;
112 }
113 
114 int main(){
115     int m;
116     while( scanf("%d%d",&n,&m)!=EOF ){
117         int a,b,c;
118         init();
119         for( int i=0;i<m;i++ ){
120             scanf("%d%d%d",&a,&b,&c);
121             addedge( a,b,c );
122         }
123         int s,t,k;
124         scanf("%d%d%d",&s,&t,&k);
125         if( s==t )k++;
126         Dij( t,s );
127         printf("%d\n",A_star( s,t,k ));
128     }
129     return 0;
130 }

 

posted @ 2013-02-23 21:07  xxx0624  阅读(737)  评论(0)  编辑  收藏  举报