HDU 1116 play on words

Problem address: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1116

 

A question about Euler's loop. I have just been in contact with oocl, so I have collected some information:

Definition of European Union:

If such a path exists in graph G so that it passes through every edge of graph G once, the path is called Euler's path. If the path is a circle, it is called Euler's (Euler) loop. An image with an Euler's path is called a map E ).

Methods to determine whether there is Euler's path:

Directed Graph: Graph connectivity. There is a vertex degree of exit greater than 1, there is a vertex degree of entry greater than 1, and the rest are degrees of exit = inbound.

Undirected graph: Graph connectivity. Only two vertices are odd, and the rest are even.

Methods To determine whether the Euler loop exists:

Directed Graph: Graph connectivity. All vertices are outdegree = inbound.

Undirected graph: Graph connectivity. All vertices are even. Reference: http://www.cnblogs.com/buptLizer/archive/2012/04/15/2450297.html http://www.cnblogs.com/zhourongqing/archive/2012/07/10/2585235.html

# Include <iostream> # include <cstring> # include <string> # include <cstdio> # include <queue> using namespace STD; const int maxn = 30; int in [maxn]; // corresponding to the inbound int out [maxn]; // corresponding to the outbound int used [maxn]; bool vis [maxn]; int find (int n) {If (n! = Used [N]) used [N] = find (used [N]); Return used [N];} void Merge (int A, int B) {// merge int x = find (a); int y = find (B); If (X! = Y) used [y] = x; return;} int main () {int t, n; int I, Pos; int injudge, outjudge; bool bug; string temp; cin> T; while (t --) {CIN> N; for (I = 0; I <maxn; I ++) used [I] = I; memset (in, 0, sizeof (in); memset (Out, 0, sizeof (out); memset (VIS, false, sizeof (VIS )); // initialize while (n --) {CIN> temp; Pos = temp. length ()-1; out [temp [0]-'a'] ++; in [temp [POS]-'a'] ++; merge (temp [0]-'A', temp [POS]-' A'); vis [temp [0]-'a'] = vis [temp [POS]-'a'] = true;} int counter = 0; // used to determine whether there are several graphs. If there are more than one graph, it cannot meet the question requirements for (I = 0; I <= 'Z'-'A'; I ++) if (vis [I] & amp; used [I] = I) Counter ++; If (counter> 1) {cout <"The door cannot be opened. \ n "; continue;} injudge = outjudge = 0; bug = false; for (I = 0; I <= 'Z'-'A'; I ++) if (vis [I]) {If (in [I] = out [I]) continue; else if (in [I]-out [I] = 1) injudge ++; // The number of incoming values greater than the outgoing value. 1 e LSE if (out [I]-in [I] = 1) outjudge ++; // Number of outgoing records greater than inbound: 1 else bug = true ;} if (Bug) {cout <"The door cannot be opened. \ n "; continue;} If (injudge = 1 & outjudge = 1) {cout <" ordering is possible. \ n "; continue;} If (! Injudge &&! Outjudge) {cout <"ordering is possible. \ n"; continue;} cout <"The door cannot be opened. \ n";} return 0 ;}

 

 

HDU 1116 play on words