poj1789

题意不好理解，其实是最小生成树。

View Code

   
     

    
      #include 
    
      <
    
      iostream
    
      >
    
      
#include 
    
      <
    
      cstdio
    
      >
    
      
#include 
    
      <
    
      cstdlib
    
      >
    
      
#include 
    
      <
    
      cstring
    
      >
    
      

    
      using
    
       
    
      namespace
    
       std;


    
      #define
    
       maxn 2005
    
      

    
      #define
    
       inf 0x3f3f3f3f
    
      


    
      int
    
       n;

    
      char
    
       code[maxn][
    
      10
    
      ];

    
      int
    
       cost[maxn][maxn];

    
      int
    
       vis[maxn];

    
      int
    
       lowc[maxn];


    
      int
    
       cal(
    
      int
    
       a, 
    
      int
    
       b)
{
 
    
      int
    
       ret 
    
      =
    
       
    
      0
    
      ;
 
    
      for
    
       (
    
      int
    
       i 
    
      =
    
       
    
      0
    
      ; i 
    
      <
    
       
    
      7
    
      ; i
    
      ++
    
      )
 
    
      if
    
       (code[a][i] 
    
      !=
    
       code[b][i])
 ret
    
      ++
    
      ;
 
    
      return
    
       ret;
}


    
      int
    
       prim()
{
 
    
      int
    
       i, j, p;
 
    
      int
    
       minc, res 
    
      =
    
       
    
      0
    
      ;
 memset(vis, 
    
      0
    
      , 
    
      sizeof
    
      (vis));
 vis[
    
      0
    
      ] 
    
      =
    
       
    
      1
    
      ;
 
    
      for
    
       (i 
    
      =
    
       
    
      1
    
      ; i 
    
      <
    
       n; i
    
      ++
    
      )
 lowc[i] 
    
      =
    
       cost[
    
      0
    
      ][i];
 
    
      for
    
       (i 
    
      =
    
       
    
      1
    
      ; i 
    
      <
    
       n; i
    
      ++
    
      )
 {
 minc 
    
      =
    
       inf;
 p 
    
      =
    
       
    
      -
    
      1
    
      ;
 
    
      for
    
       (j 
    
      =
    
       
    
      0
    
      ; j 
    
      <
    
       n; j
    
      ++
    
      )
 
    
      if
    
       (
    
      0
    
       
    
      ==
    
       vis[j] 
    
      &&
    
       minc 
    
      >
    
       lowc[j])
 {
 minc 
    
      =
    
       lowc[j];
 p 
    
      =
    
       j;
 }
 
    
      if
    
       (inf 
    
      ==
    
       minc) 
    
      return
    
       
    
      -
    
      1
    
      ;
 res 
    
      +=
    
       minc; vis[p] 
    
      =
    
       
    
      1
    
      ;
 
    
      for
    
       (j 
    
      =
    
       
    
      0
    
      ; j 
    
      <
    
       n; j
    
      ++
    
      )
 
    
      if
    
       (
    
      0
    
       
    
      ==
    
       vis[j] 
    
      &&
    
       lowc[j] 
    
      >
    
       cost[p][j])
 lowc[j] 
    
      =
    
       cost[p][j];
 }
 
    
      return
    
       res;
}


    
      int
    
       main()
{
 
    
      //
    
      freopen("t.txt", "r", stdin);
    
      

    
       
    
      while
    
       (scanf(
    
      "
    
      %d
    
      "
    
      , 
    
      &
    
      n), n)
 {
 memset(cost, 
    
      0
    
      , 
    
      sizeof
    
      (cost));
 getchar();
 
    
      for
    
       (
    
      int
    
       i 
    
      =
    
       
    
      0
    
      ; i 
    
      <
    
       n; i
    
      ++
    
      )
 gets(code[i]);
 
    
      for
    
       (
    
      int
    
       i 
    
      =
    
       
    
      0
    
      ; i 
    
      <
    
       n 
    
      -
    
       
    
      1
    
      ; i
    
      ++
    
      )
 
    
      for
    
       (
    
      int
    
       j 
    
      =
    
       i 
    
      +
    
       
    
      1
    
      ; j 
    
      <
    
       n; j
    
      ++
    
      )
 {
 cost[i][j] 
    
      =
    
       cal(i, j);
 cost[j][i] 
    
      =
    
       cost[i][j];
 }
 printf(
    
      "
    
      The highest possible quality is 1/%d.\n
    
      "
    
      , prim());
 }
 
    
      return
    
       
    
      0
    
      ;
}