poj 1789 Truck History

 where the sum goes over all pairs of types in the derivation plan such that t_o is the original type and t_d the type derived from it and d(t_o,t_d) is the distance of the types

1/Σ_(to,td)d(t_o,t_d)

派生关系有向边，边权为两个串的不同字符的个数，要求的是找出一种用到所有串的派生关系，使得Q最小，（因为是1/Q），所以这题就是求最小生成树

最朴素的最小生成树AC

  
    

   
     #include
   
     <
   
     stdio.h
   
     >
   
     

   
     const
   
      
   
     int
   
      INF 
   
     =
   
      
   
     0x3f3f3f3f
   
     ;

   
     const
   
      
   
     int
   
      MAXN 
   
     =
   
      
   
     2000
   
     ;

   
     char
   
      word[MAXN][
   
     8
   
     ];

   
     int
   
      map[MAXN][MAXN];

   
     int
   
      n, lowcost[MAXN], cloest[MAXN];

   
     bool
   
      used[MAXN];
inline 
   
     int
   
      countDis(
   
     int
   
      i, 
   
     int
   
      j){
 
   
     int
   
      cnt 
   
     =
   
      
   
     0
   
     ;
 
   
     for
   
     (
   
     int
   
      k 
   
     =
   
      
   
     0
   
     ; k 
   
     <
   
      
   
     7
   
     ; 
   
     ++
   
     k){
 
   
     if
   
     (word[i][k] 
   
     !=
   
      word[j][k])
 cnt
   
     ++
   
     ;
 }
 
   
     return
   
      cnt;
}

   
     int
   
      prim(){
 
   
     int
   
      ans 
   
     =
   
      
   
     0
   
     , min, minidx;
 
   
     for
   
     (
   
     int
   
      i 
   
     =
   
      
   
     0
   
     ; i 
   
     <
   
      n; 
   
     ++
   
     i){
 lowcost[i] 
   
     =
   
      map[
   
     0
   
     ][i];
 cloest[i] 
   
     =
   
      
   
     0
   
     ;
 used[i] 
   
     =
   
      
   
     false
   
     ;
 }
 used[
   
     0
   
     ] 
   
     =
   
      
   
     true
   
     ;
 
   
     for
   
     (
   
     int
   
      i 
   
     =
   
      
   
     1
   
     ; i 
   
     <
   
      n; 
   
     ++
   
     i){
 min 
   
     =
   
      INF;
 
   
     for
   
     (
   
     int
   
      j 
   
     =
   
      
   
     0
   
     ; j 
   
     <
   
      n; 
   
     ++
   
     j){
 
   
     if
   
     (
   
     !
   
     used[j] 
   
     &&
   
      lowcost[j] 
   
     <
   
      min)
 min 
   
     =
   
      lowcost[j], minidx 
   
     =
   
      j;
 }
 used[minidx] 
   
     =
   
      
   
     true
   
     ;
 ans 
   
     +=
   
      min;
 
   
     for
   
     (
   
     int
   
      j 
   
     =
   
      
   
     0
   
     ; j 
   
     <
   
      n; 
   
     ++
   
     j){
 
   
     if
   
     (
   
     !
   
     used[j] 
   
     &&
   
      lowcost[j] 
   
     >
   
      map[minidx][j])
 lowcost[j] 
   
     =
   
      map[minidx][j], cloest[j] 
   
     =
   
      minidx;
 }
 }
 
   
     return
   
      ans;
}

   
     int
   
      main(){
 
   
     while
   
     (scanf(
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     n), n){
 
   
     for
   
     (
   
     int
   
      i 
   
     =
   
      
   
     0
   
     ; i 
   
     <
   
      n; 
   
     ++
   
     i){
 scanf(
   
     "
   
     %s
   
     "
   
     ,word[i]);
 }
 
   
     for
   
     (
   
     int
   
      i 
   
     =
   
      
   
     0
   
     ; i 
   
     <
   
      n; 
   
     ++
   
     i){
 
   
     for
   
     (
   
     int
   
      j 
   
     =
   
      
   
     0
   
     ; j 
   
     <
   
      n; 
   
     ++
   
     j){
 map[i][j] 
   
     =
   
      countDis(i, j);
 }
 }
 printf(
   
     "
   
     The highest possible quality is 1/%d.\n
   
     "
   
     ,prim());
 }
 
   
     return
   
      
   
     0
   
     ;
}