[Euler Loop] poj 1386 play on words

Question link:

Http://poj.org/problem? Id = 1386

Play on words Time limit:1000 ms   Memory limit:10000 K Total submissions:9685   Accepted:3344 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 shocould print the sentence "ordering is possible.". Otherwise, output the sentence "The door cannot be opened .". Sample Input 32acmibm3acmmalformmouse2okok Sample output The door cannot be opened.Ordering is possible.The door cannot be opened. Source Central Europe 1999

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Question meaning:

Give n words. If the last letter of A is the same as the first letter of B, it means that a and B can be connected together. Ask if all words are in a string.

Solution:

Connectivity + Euler's path

Consider only the first letter and the last letter of a word, and use the letter as a node. The first letter and the last letter of a word are connected to one side. First, check the set to determine whether it is connected, and then determine whether there is an Euler path.

Code:

//#include<CSpreadSheet.h>#include<iostream>#include<cmath>#include<cstdio>#include<sstream>#include<cstdlib>#include<string>#include<string.h>#include<cstring>#include<algorithm>#include<vector>#include<map>#include<set>#include<stack>#include<list>#include<queue>#include<ctime>#include<bitset>#include<cmath>#define eps 1e-6#define INF 0x3f3f3f3f#define PI acos(-1.0)#define ll __int64#define LL long long#define lson l,m,(rt<<1)#define rson m+1,r,(rt<<1)|1#define M 1000000007//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define Maxn 30bool vis[Maxn];int dei[Maxn],deo[Maxn],fa[Maxn],n;char sa[1100];int Find(int x){    int temp=x;    while(fa[x]!=x)        x=fa[x];    while(fa[temp]!=x)    {        int cur=fa[temp];        fa[temp]=x;        temp=cur;    }    return x;}void Unio(int x,int y){    x=Find(x),y=Find(y);    if(x!=y)        fa[x]=y;}int main(){    //freopen("in.txt","r",stdin);   //freopen("out.txt","w",stdout);   int t;   scanf("%d",&t);   while(t--)   {        scanf("%d",&n);        memset(vis,false,sizeof(vis));        memset(dei,0,sizeof(dei));        memset(deo,0,sizeof(deo));        for(int i=1;i<=26;i++)            fa[i]=i;        while(n--)        {            scanf("%s",sa+1);            int len=strlen(sa+1);            deo[sa[1]-'a'+1]++;            dei[sa[len]-'a'+1]++;            vis[sa[1]-'a'+1]=true;            vis[sa[len]-'a'+1]=true;            Unio(sa[1]-'a'+1,sa[len]-'a'+1);        }        int la=-1,nui=0,nuo=0;        bool ans=true;        for(int i=1;i<=26;i++)        {            if(vis[i])            {                if(la==-1)                    la=Find(i);                else if(la!=Find(i))                {                     //printf("->i:%d la:%d\n",i,la);                    //system("pause");                    ans=false;                    break;                }                if(dei[i]-deo[i]==1)                    nui++;                else if(deo[i]-dei[i]==1)                    nuo++;                else if(deo[i]!=dei[i])                {                    ans=false;                    break;                }                //printf("i:%d la:%d\n",i,la);                    //system("pause");            }        }        if(!ans||nui>=2||nuo>=2||(nui+nuo)==1)            printf("The door cannot be opened.\n");        else            printf("Ordering is possible.\n");   }    return 0;}

[Euler Loop] poj 1386 play on words