(Query set + Euler's path) play on words

Question:

There are n words separated by lowercase letters. It is required to determine whether there are two adjacent words in a certain arrangement. The last letter of the first word is the same as the first letter of the last word.

Analysis:

When two letters of a word are considered as nodes, a word can be considered as a directed edge. The arrangement in the question is equivalent to the existence of Euler's path in the directed graph.

Before judging, you must first make sure that the graph is connected, and use the query set in the code.

Let's look back at the conditions with Euler's path. They are all even points or there are only two singularity. DEG is used to record the degree of each vertex. the outbound degree is 1, and the inbound degree is-1.

The condition in the program for determining the existence of Euler's path is: DEG is all 0 or there are two not 0, one of which is 1 and the other is-1.

Used records whether a letter has appeared.

 

 1 //#define LOCAL 2 #include <vector> 3 #include <cstdio> 4 #include <cstring> 5 using namespace std; 6  7 const int maxn = 1000 + 10; 8 char word[maxn]; 9 int pa[256], deg[256], cc, used[256];10 11 int find(int a)12 { return pa[a] == a ? a : pa[a] = find(pa[a]); }13 14 int main(void)15 {16     #ifdef LOCAL17         freopen("10129in.txt", "r", stdin);18     #endif19 20     int T, n;21     scanf("%d", &T);22     while(T--)23     {24         memset(used, 0, sizeof(used));25         memset(deg, 0, sizeof(deg));26         for(int i = ‘a‘; i <= ‘z‘; ++i)27             pa[i] = i;28         cc = 26;    //Á¬Í¨¿éµÄÊýÁ¿ 29 30         scanf("%d", &n);31         for(int i = 0; i < n; ++i)32         {33             scanf("%s", word);34             char c1 = word[0];35             char c2 = word[strlen(word) - 1];36             used[c1] = used[c2] = 1;37             deg[c1]++;    deg[c2]--;38             int p1 = find(c1);39             int p2 = find(c2);40             if(p1 != p2)41             {42                 cc--;43                 pa[p1] = p2;44             }45         }46 47         vector<int> d;48         for(int i = ‘a‘; i <= ‘z‘; ++i)49         {50             if(!used[i]) --cc;51             else if(deg[i])    d.push_back(i);52         }53         bool ok = false;54         if(cc == 1 && (d.empty() || (d.size() == 2 && (deg[d[0]] == 1 || deg[d[0]] == -1)))) ok = true;55         if(ok)    puts("Ordering is possible.");56         else puts("The door cannot be opened.");57     }58 59     return 0;60 }

Code Jun

 

(Query set + Euler's path) play on words