Time Limit: ?1000MS | ? | Memory Limit: ?10000K |
Total Submissions: ?8571 | ? | Accepted: ?2997 |
Description
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
Output
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 }
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。
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