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Play on Words(有向圖歐拉路)

系統(tǒng) 1573 0
Time Limit: ?1000MS ? Memory Limit: ?10000K
Total Submissions: ?8571 ? Accepted: ?2997

Description

Some of the secret doors contain a very interesting word puzzle. The team of archaeologists has to solve it to open that doors. Because there is no other way to open the doors, the puzzle is very important for us.?

There is a large number of magnetic plates on every door. Every plate has one word written on it. The plates must be arranged into a sequence in such a way that every word begins with the same letter as the previous word ends. For example, the word ``acm'' can be followed by the word ``motorola''. Your task is to write a computer program that will read the list of words and determine whether it is possible to arrange all of the plates in a sequence (according to the given rule) and consequently to open the door.?

Input

The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing a single integer number Nthat indicates the number of plates (1 <= N <= 100000). Then exactly Nlines follow, each containing a single word. Each word contains at least two and at most 1000 lowercase characters, that means only letters 'a' through 'z' will appear in the word. The same word may appear several times in the list.

Output

Your program has to determine whether it is possible to arrange all the plates in a sequence such that the first letter of each word is equal to the last letter of the previous word. All the plates from the list must be used, each exactly once. The words mentioned several times must be used that number of times.?
If there exists such an ordering of plates, your program should print the sentence "Ordering is possible.". Otherwise, output the sentence "The door cannot be opened.".?

Sample Input

    3

2

acm

ibm

3

acm

malform

mouse

2

ok

ok


  

Sample Output

    The door cannot be opened.

Ordering is possible.

The door cannot be opened.
    

題意:給出n個(gè)單詞,問(wèn)所有的單詞能否首尾相連(能相連的單詞的首和尾必須是相同的);

思路:一道判斷有向圖歐拉路的題目;
   可以將每個(gè)單詞的首和尾看作節(jié)點(diǎn),判斷圖的連通性可以用并查集,每輸入一個(gè)單詞將其首和尾加入集合中,最后任取一個(gè)節(jié)點(diǎn)判斷其他所有節(jié)點(diǎn)和該節(jié)點(diǎn)是否有共同的祖先,若是,則是連通的,否則不連通;
         在連通性的前提下,若所有節(jié)點(diǎn)的入讀等于出度 或者 一個(gè)節(jié)點(diǎn)的入度比出度大1 一個(gè)節(jié)點(diǎn)的入度比出度小1,說(shuō)明有歐拉路,否則沒(méi)有歐拉路;
    
因?yàn)槭终`,貢獻(xiàn)一次WA;
        
            1
        
         #include<stdio.h>


        
            2
        
         #include<
        
          string
        
        .h>


        
            3
        
        
            4
        
        
          int
        
        
          set
        
        [
        
          30
        
        ],indegree[
        
          1010
        
        ],outdegree[
        
          1010
        
        ],vis[
        
          26
        
        
          ];


        
        
            5
        
        
          int
        
        
           count;


        
        
            6
        
        
            7
        
        
          void
        
        
           init()


        
        
            8
        
        
          {


        
        
            9
        
        
          for
        
        (
        
          int
        
         i = 
        
          0
        
        ; i < 
        
          26
        
        ; i++
        
          )


        
        
           10
        
        
          set
        
        [i] =
        
           i;


        
        
           11
        
        
          }


        
        
           12
        
        
           13
        
        
          int
        
         find(
        
          int
        
        
           x)


        
        
           14
        
        
          {


        
        
           15
        
        
          if
        
        (
        
          set
        
        [x] !=
        
           x)


        
        
           16
        
        
          set
        
        [x] = find(
        
          set
        
        
          [x]);


        
        
           17
        
        
          return
        
        
          set
        
        
          [x];


        
        
           18
        
        
          }


        
        
           19
        
        
           20
        
        
          int
        
        
           check()


        
        
           21
        
        
          {


        
        
           22
        
        
          int
        
        
           x,i;


        
        
           23
        
        
          int
        
         flag = 
        
          0
        
        
          ;


        
        
           24
        
        
          for
        
        (i = 
        
          0
        
        ; i < 
        
          26
        
        ; i++
        
          )


        
        
           25
        
        
              {


        
        
           26
        
        
          if
        
        
          (vis[i])


        
        
           27
        
        
                  {


        
        
           28
        
        
          if
        
        (flag == 
        
          0
        
        
          )


        
        
           29
        
        
                      {


        
        
           30
        
                         x =
        
           find(i);


        
        
           31
        
                         flag = 
        
          1
        
        
          ;


        
        
           32
        
        
                      }


        
        
           33
        
        
          else
        
        
           34
        
        
                      {


        
        
           35
        
        
          if
        
        (find(i) !=
        
           x)


        
        
           36
        
        
          break
        
        
          ;


        
        
           37
        
        
                      }


        
        
           38
        
        
                  }


        
        
           39
        
        
              }


        
        
           40
        
        
          if
        
        (i < 
        
          26
        
        
          )


        
        
           41
        
        
          return
        
        
          0
        
        ;
        
          //
        
        
          圖是不連通的,直接返回;
        
        
           42
        
        
           43
        
        
          int
        
         c1 = 
        
          0
        
        , c2 = 
        
          0
        
        , c3 = 
        
          0
        
        
          ;


        
        
           44
        
        
          for
        
        (
        
          int
        
         i = 
        
          0
        
        ; i < 
        
          26
        
        ; i++
        
          )


        
        
           45
        
        
              {


        
        
           46
        
        
          if
        
        
          (vis[i])


        
        
           47
        
        
                  {


        
        
           48
        
        
          if
        
        (indegree[i] ==
        
           outdegree[i])


        
        
           49
        
                         c1++
        
          ;


        
        
           50
        
        
          else
        
        
          if
        
        (indegree[i]-outdegree[i] == 
        
          1
        
        
          )


        
        
           51
        
                         c2++
        
          ;


        
        
           52
        
        
          else
        
        
          if
        
        (outdegree[i]-indegree[i] == 
        
          1
        
        
          )


        
        
           53
        
                         c3++
        
          ;


        
        
           54
        
        
                  }


        
        
           55
        
        
              }


        
        
           56
        
        
          if
        
        ((c2 == 
        
          1
        
         && c3 == 
        
          1
        
         && c1 == count-
        
          2
        
        ) ||(c1 ==
        
           count))


        
        
           57
        
        
          return
        
        
          1
        
        
          ;


        
        
           58
        
        
          else
        
        
          return
        
        
          0
        
        
          ;


        
        
           59
        
        
          }


        
        
           60
        
        
           61
        
        
          int
        
        
           main()


        
        
           62
        
        
          {


        
        
           63
        
        
          int
        
        
           test,n;


        
        
           64
        
        
          char
        
         s[
        
          1010
        
        
          ];


        
        
           65
        
             scanf(
        
          "
        
        
          %d
        
        
          "
        
        ,&
        
          test);


        
        
           66
        
        
          while
        
        (test--
        
          )


        
        
           67
        
        
              {


        
        
           68
        
                 memset(indegree,
        
          0
        
        ,
        
          sizeof
        
        
          (indegree));


        
        
           69
        
                 memset(outdegree,
        
          0
        
        ,
        
          sizeof
        
        
          (outdegree));


        
        
           70
        
                 memset(vis,
        
          0
        
        ,
        
          sizeof
        
        
          (vis));


        
        
           71
        
        
                  init();


        
        
           72
        
                 count = 
        
          0
        
        
          ;


        
        
           73
        
        
           74
        
                 scanf(
        
          "
        
        
          %d
        
        
          "
        
        ,&
        
          n);


        
        
           75
        
        
          for
        
        (
        
          int
        
         i = 
        
          1
        
        ; i <= n; i++
        
          )


        
        
           76
        
        
                  {


        
        
           77
        
                     scanf(
        
          "
        
        
          %s
        
        
          "
        
        
          ,s);


        
        
           78
        
        
          int
        
         len =
        
           strlen(s);


        
        
           79
        
        
          int
        
         u = s[
        
          0
        
        ]-
        
          '
        
        
          a
        
        
          '
        
        
          ;


        
        
           80
        
        
          if
        
        (!
        
          vis[u])


        
        
           81
        
        
                      {


        
        
           82
        
                         vis[u] = 
        
          1
        
        
          ;


        
        
           83
        
                         count++
        
          ;


        
        
           84
        
        
                      }


        
        
           85
        
        
          int
        
         v = s[len-
        
          1
        
        ]-
        
          '
        
        
          a
        
        
          '
        
        
          ;


        
        
           86
        
        
          if
        
        (!
        
          vis[v])


        
        
           87
        
        
                      {


        
        
           88
        
                         vis[v] = 
        
          1
        
        
          ;


        
        
           89
        
                         count++
        
          ;


        
        
           90
        
        
                      }


        
        
           91
        
        
           92
        
                     indegree[u]++
        
          ;


        
        
           93
        
                     outdegree[v]++
        
          ;


        
        
           94
        
        
          int
        
         tu =
        
           find(u);


        
        
           95
        
        
          int
        
         tv =
        
           find(v);


        
        
           96
        
        
          if
        
        (tu !=
        
           tv)


        
        
           97
        
        
          set
        
        [tu] =
        
           tv;


        
        
           98
        
        
                  }


        
        
           99
        
        
          100
        
        
          if
        
        
          (check())


        
        
          101
        
                     printf(
        
          "
        
        
          Ordering is possible.\n
        
        
          "
        
        
          );


        
        
          102
        
        
          else
        
         printf(
        
          "
        
        
          The door cannot be opened.\n
        
        
          "
        
        
          );


        
        
          103
        
        
              }


        
        
          104
        
        
          return
        
        
          0
        
        
          ;


        
        
          105
        
         }
      
View Code
  

歐拉路 圖G,若存在一條路,經(jīng)過(guò)G中每條邊有且僅有一次,稱(chēng)這條路為歐拉 路, 如果存在一條回路經(jīng)過(guò)G每條邊有且僅有一次,

稱(chēng)這條回路為歐拉回路。具有歐拉回路的圖成為歐拉圖。

?

判斷 歐拉路 是否存在的方法

有向圖 :圖連通,有一個(gè)頂點(diǎn)出度大入度1,有一個(gè)頂點(diǎn)入度大出度1,其余都是出度=入度。

無(wú)向圖 :圖連通,只有兩個(gè)頂點(diǎn)是奇數(shù)度,其余都是偶數(shù)度的。

?

判斷 歐拉回路 是否存在的方法

有向圖 :圖連通,所有的頂點(diǎn)出度=入度。

無(wú)向圖 :圖連通,所有頂點(diǎn)都是偶數(shù)度。

其中判斷圖的連通性用并查集。

Play on Words(有向圖歐拉路)


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