zoj2016Play on words(유향도 유라 회로 통로 존재 여부 판단)

Play on Words
Time Limit: 5 Seconds     
Memory Limit: 32768 KB 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. 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.
제목은 다음과 같다. 한 조의 단어를 제시하여 한 조의 배열을 찾을 수 있는지를 보고 이 조의 배열 중의 이전 단어의 마지막 자모가 다음 단어의 첫 번째 자모이도록 한다.
각 단어는 머리와 꼬리 두 자모를 연결하는 한 변으로 볼 수 있으며, 이렇게 하면 머리와 꼬리 각 단어가 하는 점의 출도와 입도를 통계할 수 있다.
1. 각 노드의 출입도가 같으면 오라회로가 있을 수 있다.
2. 두 노드만 출입도가 다르고 출입도 차이가 1과 -1이면 오라통로가 있다.
다른 상황은 성립될 가능성이 없다.

#include<stdio.h>
#include<string.h>
struct edge{
    int u,v;
}edges[100005];
int out[26],in[26];
int used[26];
int n;
int parent[26];
char str[1005];
void UFset()
{
    for(int i=0;i<26;i++)
    parent[i]=-1;
}
int Find(int x)
{
    int s;
    for(s=x;parent[s]>=0;s=parent[s]);
    while(s!=x)
    {
        int temp=parent[x];
        parent[x]=s;
        x=temp;
    }
    return s;
}
void Union(int x1,int x2)
{
    int r1=Find(x1),r2=Find(x2);
    int temp=parent[r1]+parent[r2];
    if(parent[r1]<parent[r2])
    {
        parent[r2]=r1;
        parent[r1]=temp;
    }
    else
    {
        parent[r1]=r2;
        parent[r2]=temp;
    }
}
bool isconnect()
{
    int u,v,i;
    UFset();
    for(i=0;i<n;i++)
    {
        u=edges[i].u;v=edges[i].v;
        if(u!=v&&Find(u)!=Find(v))
        Union(u,v);
    }
    int first=-1;
    for(i=0;i<26;i++)
    {
        if(!used[i]) continue;
        if(first==-1) first=i;
        else if(Find(i)!=Find(first)) break;
    }
    if(i<26) return false;
    else return true;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(out,0,sizeof(out));
        memset(in,0,sizeof(in));
        memset(used,0,sizeof(used));
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%s",str);
            edges[i].u=str[0]-'a';  edges[i].v=str[strlen(str)-1]-'a';
            out[str[0]-'a']++;  in[str[strlen(str)-1]-'a']++;
            used[str[0]-'a']=1; used[str[strlen(str)-1]-'a']=1;
        }
        int judge1=0,judge2=0;
        bool flag1=true;
        for(int i=0;i<26;i++)
        {
            if(!used[i]) continue;
            if(out[i]-in[i]>=2||in[i]-out[i]>=2) {flag1=false;break;}
            if(out[i]==0&&in[i]==0) {flag1=false;break;}
            if(out[i]-in[i]==1)
            {  judge1++;
               if(judge1>1) {flag1=false;break;}
            }
            if(in[i]-out[i]==1)
            {  judge2++;
               if(judge2>1) {flag1=false;break;}
            }
        }
        if(judge1!=judge2) flag1=false;
        if(flag1==true&&isconnect())
        printf("Ordering is possible.
");
        else
        printf("The door cannot be opened.
");
    }
    return 0;
}