POJ2449 A*+第K短路

题意：s到t的第K短路

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  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 }