POJ - 1386 Play on Words(基础欧拉回路)

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Play on Words Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12369   Accepted: 4198 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 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. Source Central Europe 1999

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基础欧拉回路，词语接龙，只需要存首尾字母即可。

首先并查集判断是否只有一个连通集，然后判断欧拉路径的条件：除首尾点外，其他点入度==出度。首点入度比出度大一，尾点出度比入度大一。

附上AC代码：

#include
#include
#include

using namespace std;
const int maxn=30;
int in[maxn],out[maxn];
int vis[maxn];
int T,N;
int innum,outnum,numpar,oddnum;
char c[1000+5];
int par[maxn];

void init()
{
    for(int i=1;i<=30;i++)
        par[i]=i;
        memset(in,0,sizeof(in));
        memset(out,0,sizeof(out));
        memset(vis,0,sizeof(vis));
        innum=0;outnum=0;numpar=0;
}

int find(int a)
{
    if(a==par[a])return a;
    return par[a]=find(par[a]);
}

void unite(int a,int b)
{
    a=find(a);
    b=find(b);
    if(a==b)return ;
    else if(a>b)
        par[a]=b;
    else
        par[b]=a;
}

int main()
{
    ios::sync_with_stdio(false);
    cin>>T;
    while(T--)
    {
        init();
        cin>>N;
        for(int i=0;i>c;
            int len=strlen(c);
            in[c[len-1]-'a'+1]++;
            out[c[0]-'a'+1]++;
            unite(c[0]-'a'+1,c[len-1]-'a'+1);
            vis[c[0]-'a'+1]=1;
            vis[c[len-1]-'a'+1]=1;
        }
        int flag=1;
        for(int i=1;i<30;i++)
        {
            if(vis[i])
            {
                if(par[i]==i)numpar++;
                if(in[i]!=out[i])
                {
                    if(in[i]-out[i]==1)innum++;
                    else if(out[i]-in[i]==1)outnum++;
                    else {flag=0;}
                }
                if(numpar>1){flag=0;break;}
            }
        }
        if((flag&&innum==0&&outnum==0)||(flag&&innum==1&&outnum==1))
            cout<<"Ordering is possible."<