POJ 1125 Stockbroker Grapevine

给你n个人的联系情况，对任意一个人，求出这个人发消息到其他n-1个人的时间，得到n-1个时间中的最大值，n个最大值中的最小值就是所求。如果网络不通，那就输出disjoint

Floyed算出任意两个人的最小时间 就OK了

 
   

  
    
   
     

    
      #include 
    
      <
    
      iostream
    
      >
    
      
#include 
    
      <
    
      fstream
    
      >
    
      

    
      using
    
       
    
      namespace
    
       std;

    
      const
    
       
    
      int
    
       MAXN 
    
      =
    
       
    
      101
    
      ;

    
      const
    
       
    
      int
    
       INF 
    
      =
    
       
    
      0x7FFF
    
      ;

    
      int
    
       m , n;

    
      int
    
       G[MAXN][MAXN];


    
      void
    
       
    
      in
    
      (){
 
    
      int
    
       i,j,w,t;
 
    
      for
    
      (i 
    
      =
    
       
    
      1
    
      ; i 
    
      <=
    
       m;i
    
      ++
    
      )
 
    
      for
    
      (j 
    
      =
    
       
    
      1
    
      ; j 
    
      <=
    
       m; j
    
      ++
    
      )
 G[i][j] 
    
      =
    
       INF;
 
    
      for
    
      (i 
    
      =
    
       
    
      1
    
      ; i 
    
      <=
    
       m; i
    
      ++
    
      )
 G[i][i] 
    
      =
    
       
    
      0
    
      ;
 
    
      for
    
      (i 
    
      =
    
       
    
      1
    
      ; i 
    
      <=
    
       m; i
    
      ++
    
      ){
 cin 
    
      >>
    
       n;
 
    
      for
    
      (j 
    
      =
    
       
    
      1
    
      ; j 
    
      <=
    
       n;j
    
      ++
    
      ){
 cin 
    
      >>
    
       w 
    
      >>
    
       t;
 G[i][w] 
    
      =
    
       t;
 }
 }
}


    
      void
    
       floyd(){
 
    
      int
    
       i,j,k;
 
    
      for
    
      (k 
    
      =
    
       
    
      1
    
      ; k 
    
      <=
    
       m; k
    
      ++
    
      )
 
    
      for
    
      (i 
    
      =
    
       
    
      1
    
      ; i 
    
      <=
    
       m; i
    
      ++
    
      )
 
    
      for
    
      (j 
    
      =
    
       
    
      1
    
      ; j 
    
      <=
    
       m; j
    
      ++
    
      ){
 
    
      if
    
      (G[i][j] 
    
      >
    
       G[i][k] 
    
      +
    
       G[k][j])
 G[i][j] 
    
      =
    
       G[i][k] 
    
      +
    
       G[k][j];
 }

}


    
      void
    
       
    
      out
    
      (){
 
    
      int
    
       i,j,temp 
    
      =
    
       
    
      0
    
      ,bestN,best 
    
      =
    
       INF;
 
    
      bool
    
       flag 
    
      =
    
       
    
      false
    
      ;
 
    
      for
    
      (i 
    
      =
    
       
    
      1
    
      ; i 
    
      <=
    
       m; i
    
      ++
    
      ){
 flag 
    
      =
    
       
    
      false
    
      ;
 temp 
    
      =
    
       
    
      0
    
      ;
 
    
      for
    
      (j 
    
      =
    
       
    
      1
    
      ; j 
    
      <=
    
       m; j
    
      ++
    
      ){
 
    
      if
    
      ( temp 
    
      <
    
       G[i][j] 
    
      &&
    
       i 
    
      !=
    
       j){
 temp 
    
      =
    
       G[i][j];
 }
 
    
      if
    
      (G[i][j] 
    
      ==
    
       INF){
 flag 
    
      =
    
       
    
      true
    
      ;
 
    
      break
    
      ;
 }
 }
 
    
      if
    
      (
    
      !
    
      flag){
 
    
      if
    
      (best 
    
      >
    
       temp ){
 best 
    
      =
    
       temp;
 bestN 
    
      =
    
       i;
 } 
 }
 }
 
    
      if
    
      (bestN 
    
      ==
    
       INF)
 cout 
    
      <<
    
       
    
      "
    
      disjoint
    
      "
    
       
    
      <<
    
       endl;
 
    
      else
    
      
 cout 
    
      <<
    
       bestN 
    
      <<
    
       
    
      "
    
       
    
      "
    
       
    
      <<
    
       best 
    
      <<
    
       endl;
}


    
      int
    
       main(){
 
    
      while
    
      (cin 
    
      >>
    
       m, m
    
      !=
    
      0
    
      ){
 
    
      in
    
      ();
 floyd();
 
    
      out
    
      ();
 }
 
    
      return
    
       
    
      0
    
      ;
}