Truck History--POJ 1789

1、题目类型：图论、最小生成树、Prim算法。

2、解题思路：（1）将输入转换为map[][]：即匹配任意两个字符串，记录其不同的字母个数；（2）Prim算法求解最小生成树。

3、注意事项：Prim算法的简单应用。

4、实现方法:

  
    

   
     
   
     #include
   
     <
   
     iostream
   
     >
   
     
#include
   
     <
   
     algorithm
   
     >
   
     

   
     using
   
      
   
     namespace
   
      std;

   
     #define
   
      Max 100000
   
     


   
     int
   
      n,map[
   
     2010
   
     ][
   
     2010
   
     ];

   
     bool
   
      vis[
   
     2010
   
     ];

   
     char
   
      str[
   
     2010
   
     ][
   
     8
   
     ];


   
     int
   
      Distance(
   
     char
   
      str1[],
   
     char
   
      str2[])
{
 
   
     int
   
      i,cnt
   
     =
   
     0
   
     ;
 
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     8
   
     ;i
   
     ++
   
     )
 {
 
   
     if
   
     (str1[i]
   
     !=
   
     str2[i])
 cnt
   
     ++
   
     ;
 }
 
   
     return
   
      cnt;
}


   
     int
   
      Prim()
{
 
   
     int
   
      i,j,pos,ans
   
     =
   
     0
   
     ,dis[
   
     2010
   
     ];
 memset(vis,
   
     0
   
     ,
   
     sizeof
   
     (vis));

 
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )
 dis[i]
   
     =
   
     map[
   
     0
   
     ][i];
 vis[
   
     0
   
     ]
   
     =
   
     1
   
     ;

 
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     n
   
     -
   
     1
   
     ;i
   
     ++
   
     )
 {
 pos
   
     =
   
     min_element(dis
   
     +
   
     1
   
     ,dis
   
     +
   
     n)
   
     -
   
     dis;
 vis[pos]
   
     =
   
     1
   
     ;
 ans
   
     +=
   
     dis[pos];
 dis[pos]
   
     =
   
     Max;
 
 
   
     for
   
     (j
   
     =
   
     0
   
     ;j
   
     <
   
     n;j
   
     ++
   
     )
 
   
     if
   
     (
   
     !
   
     vis[j] 
   
     &&
   
      dis[j]
   
     >
   
     map[pos][j])
 dis[j]
   
     =
   
     map[pos][j];
 }
 
   
     return
   
      ans;
}


   
     int
   
      main()
{
 
   
     int
   
      i,j,ans;
 
   
     while
   
     (cin
   
     >>
   
     n 
   
     &&
   
      n)
 {
 
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )
 {
 cin
   
     >>
   
     str[i];
 
   
     for
   
     (j
   
     =
   
     i
   
     -
   
     1
   
     ;j
   
     >=
   
     0
   
     ;j
   
     --
   
     )
 map[i][j]
   
     =
   
     map[j][i]
   
     =
   
     Distance(str[i],str[j]);
 }
 ans
   
     =
   
     Prim();
 cout
   
     <<
   
     "
   
     The highest possible quality is 1/
   
     "
   
     <<
   
     ans
   
     <<
   
     "
   
     .
   
     "
   
     <<
   
     endl;
 }
 
   
     return
   
      
   
     1
   
     ;
}