Acm hdu 3665 seaside

Seaside

Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 396 accepted submission (s): 272

Problem descriptionxiaoy is living in a big city, there are n towns in it and some towns near the sea. all these towns are numbered from 0 to N-1 and xiaoy lives in the town numbered '0 '. there are some directed roads connecting them. it is guaranteed that you can reach any town from the town numbered '0', but not all towns connect to each other by roads directly, and there is no ring in this city. one day, xiaoy want to go to the seaside, he asks you to help him find out the shortest way.

Inputthere are several test cases. in each cases the first line contains an integer N (0 <= n <= 10), indicating the number of the towns. then followed n blocks of data, in block-I there are two integers, mi (0 <= mi <= N-1) and PI, then MI lines followed. mi means there are Mi roads beginning with the I-th Town. pi indicates whether the I-th Town is near to the sea, Pi = 0 means no, Pi = 1 means yes. in next mi lines, each line contains two integers SMI and LMIs, which means that the distance between the I-th Town and the SMI town is LMIS.

Outputeach case takes one line, print the shortest length that xiaoy reach seaside.

Sample input5
1 0
1 1
2 0
2 3
3 1
1 1
4 100
0 1
0 1

Sample output2

Source2010 Asia Regional Harbin

Recommendlcy

 # Include  <  Stdio. h  >  
# Include  <  Iostream  >  
  Using     Namespace STD;
  Const     Int  INF  =  65533  ;
  Const     Int  Maxn  =  15  ;
  Int  Distances [maxn] [maxn];
 Int  A [maxn];  //  Whether the record is near the beach  
  Int  N;
  Void  Floyed ()
{
  Int  I, J, K;
  For  (K  =  0  ; K  < N; k  ++  )
  For  (I  =  0  ; I  <  N; I  ++  )
  For  (J  =  0  ; J  < N; j  ++  )
{
  If  (Distances [I] [J]  >  Distances [I] [k]  +  Distances [k] [J])
Distances [I] [J]  =  Distances [I] [k]  +  Distances [k] [J];
}
}
  Int  Main ()
{
 //  Freopen ("test. In", "r", stdin );
  //  Freopen ("test. Out", "W", stdout );  
    Int  M;
  Int  CNT;
  Int  I, J;
  Int  B;
  Int  DIS;
 While  (Scanf (  "  % D  "  ,  &  N)  ! =  EOF)
{
  For  (I  =  0  ; I  <  N; I  ++ )
  For  (J  =  0  ; J  <  N; j  ++  )
{
  If  (I  =  J) distances [I] [J]  =  0  ;
  Else 
Distances [I] [J]  =  INF;
}
  For  (I  =  0  ; I  <  N; I  ++  )
{
Scanf (  "  % D  "  ,  & M,  &  A [I]);
  For  (J  =  0  ; J  <  M; j  ++  )
{
Scanf (  "  % D  "  ,  & B,  &  Dis );
Distances [I] [B]  =  DIS;
}
}
Floyed ();
CNT  =  INF;
  For  (I  =  0  ; I  <  N; I  ++  )
 If  (A [I])
  If  (Distances [  0  ] [I]  <  CNT) CNT  =  Distances [  0  ] [I];
Printf (  "  % D \ n  "  , CNT );

}
  Return    0  ;
}