UVA 10129 Play on Words (Oraton Road)

Test instructions

　Enter n (n <= 100000) words, whether all these words can be in a sequence, so that the first letter of each word is the same as the last letter of the previous word (for example: Acm,malform,mouse). Each word contains a maximum of 1000 lowercase letters. You can have duplicate words in the input.

Ideas:

　The two ends of a letter are opened to a node, the word as a forward edge, if the problem is borrowed, when and only if there is Oraton road in the diagram. All that is needed to determine whether the graph constructed by the word exists Oraton road, because there is a forward edge, so the use of the direction of the Tuola access to the decision can be.

Judging conditions

(1): Basemap is connected graph

(2): There can be two singularity, one of the degree of the ratio of 1, and the other one into the degree of a greater than 1.

For condition 1, here with and check set judgment, condition 2 statistics each point of the degree, into the degree, to judge on the line.

Code:

#include <stdio.h>#include<string.h>#include<iostream>#include<math.h>#include<queue>#include<stack>#include<algorithm>using namespacestd; Const intMAXV = in;intm;intPRE[MAXV +7];intOUTDEGREE[MAXV +7];intINDEGREE[MAXV +7];intFind (intx) {returnx = = Pre[x]? X:PRE[X] =Find (pre[x]);} And look up the search setvoidInitpre () { for(inti =0; I <= MAXV; i++) Pre[i] =i;} Initialize and check the array of the setintMixintXinty)//and merge of the check set {intFX = Find (x), FY =Find (y); if(FX! = FY) Pre[fx] =fy; }BOOLisconnct ()//Determine whether the graph is connected, that is, all the points are in a set {intCNT =0;  for(inti =1; I <= MAXV; i++)if((outdegree[i]! =0|| Indegree[i]! =0) && Pre[i] = = i) cnt++; if(CNT = =1)return true; return false;}BOOLIseulur ()//whether there is Euler pathway {intCNT =0; intFlag =0;  for(inti =1; I <= -; i++)        if((outdegree[i]! =0|| Indegree[i]! =0) && (indegree[i]! =Outdegree[i]))        Judging the singularity, the method is not unique. {CNT++; Flag+ = (Indegree[i]-Outdegree[i]); if(Flag >1|| Flag <-1)return false; }     if(CNT = =0|| CNT = =2&& flag = =0)return true; return false;}intMain () {intT; scanf ("%d", &T);  while(t--) {scanf ("%d", &m);        Initpre (); memset (Indegree,0,sizeof(Indegree)); memset (Outdegree,0,sizeof(Outdegree));  for(inti =1; I <= m; i++)        {            Charword[ ++7]; scanf ("%s", Word); intU = word[0] -'a'+1; intLen =strlen (word); intv = word[len-1] -'a'+1;            Mix (U, v); ++Outdegree[u]; ++Indegree[v]; }        if(Iseulur () && isconnct ()) printf ("Ordering is possible.\n"); Elseprintf"The door cannot be opened.\n"); }    return 0;}

UVA 10129 Play on Words (Oraton Road)