POJ1386Play on Words（欧拉回路）

                                                         Play on Words

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11748		Accepted: 4018

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.
【题意】这个就相当于成语接龙，需满足前一个单词的尾字母与后一个的首字母相同。然后就是问是否 存在欧拉通路。
【分析】用两个数组存每个字母的入度与出度，再用vis[]判断字母是否出现，然后并查集判断是否联通。
定理1：无向图G存在欧拉通路的条件是：G为连通图，并且G只有两个奇度节点或者无奇度节点。
推论1：（1）当G是仅有两个奇度节点的连通图时，G的欧拉通路必以此两个节点为端点。
      （2）当G是无奇度节点连通图时，G比为欧拉回路。
      （3）G为欧拉图（存在欧拉回路）的充要条件是G为无奇度节点的连通图。
定理2：有向图D存在欧拉通路的充要条件是：D为有向图，D的基图联通，并且所有顶点的出度与入度都相等；或者除两个顶点外
      其余顶点的出度与入度都相等，而这两个顶点中一个顶点的出度-入度==1，另一个出度-入度==-1；

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include<functional>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define pi acos(-1.0)
using namespace std;
typedef long long ll;
const int N=;
const int M=;
ll power(ll a,int b,ll c){ll ans=;while(b){if(b%==){ans=(ans*a)%c;b--;}b/=;a=a*a%c;}return ans;}
char str[N];
int n,m;
int vis[],pre[];
int out[],in[];
struct man
{
    int u,v;
}edg[N];
void init()
{
    memset(vis,,sizeof(vis));
    memset(out,,sizeof(out));
    memset(in,,sizeof(in));
    for(int i=;i<;i++)pre[i]=i;
}
int Find(int x) {
    if(pre[x] != x) pre[x] = Find(pre[x]);
    return pre[x];
}

void Union(int x,int y) {
    x = Find(x);
    y = Find(y);
    if(x == y) return;
    pre[y] = x;
}
bool beconnect()
{
    for(int i=;i<n;i++){
        int u=edg[i].u,v=edg[i].v;
        if(u!=v&&Find(u)!=Find(v))Union(u,v);
    }
    int f=-,k;
    for(k=;k<;k++){
        if(!vis[k])continue;
        if(f==-)f=k;
        else if(Find(k)!=Find(f))break;
    }
    if(k<)return false;
    return true;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        init();
        scanf("%d",&n);
        for(int i=;i<n;i++){
            scanf("%s",str);
            int u=str[]-'a',v=str[strlen(str)-]-'a';
            edg[i].u=u;edg[i].v=v;
            vis[u]=vis[v]=;
            out[u]++;in[v]++;
        }
        bool flag=true;
        int cnt1=,cnt2=;
        for(int i=;i<;i++){
            if(!vis[i])continue;
            if(abs(out[i]-in[i])>){
                flag=false;
                break;
            }
            if(out[i]-in[i]==){
                cnt1++;
                if(cnt1>){
                    flag=false;
                    break;
                }
            }
            if(out[i]-in[i]==-){
                cnt2++;
                if(cnt2>){
                    flag=false;
                    break;
                }
            }
        }
        if(cnt1!=cnt2)flag=false;
        if(!beconnect())flag=false;
        if(flag)puts("Ordering is possible.");
        else puts("The door cannot be opened.");
    }
    return ;
}