POJ1386Play on Words[有向图欧拉路]

Play on Words

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11846		Accepted: 4050

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

---

一些知识：

1、无向图存在欧拉回路条件：每个点度都为偶数

2、无向图存在欧拉路条件：只有两个点度为奇数或每个点度都为偶数

3、有向图存在欧拉回路条件：每个点入度都等于出度

4、有向图存在欧拉路条件：只存在这样两个点：一个点入度等于出度+1，另一个点入度=出度-1，其他每个点入度都等于出度，或者每个点入度都等于出度。

判连通，判条件就好了

奇怪的是，dfs就WA，并查集就可以

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
typedef long long ll;
const int N=1e5+5,L=1e3+5;
int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}
int T,n,ind[30],outd[30],has[30];
char s[L];
//int vis[30],g[30][30];
//void dfs(int u){//printf("dfs %d
",u);
//    vis[u]=1;
//    for(int v=0;v<=25;v++)
//        if(!vis[v]) dfs(v);
//}
int fa[30],root=0;
inline int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
int main(){
    T=read();
    while(T--){
        int flag=1;
        for(int i=0;i<=25;i++) ind[i]=outd[i]=has[i]=0,fa[i]=i;
        //memset(g,0,sizeof(g));
        n=read();
        for(int i=1;i<=n;i++){
            scanf("%s",s+1);
            int len=strlen(s+1),u=s[1]-'a',v=s[len]-'a';
            //g[u][v]=g[v][u]=1;
            fa[find(u)]=find(v);root=find(v);
            outd[u]++;ind[v]++;
            has[u]=has[v]=1;
        }
        //for(int i=0;i<=25;i++) if(has[i]) {dfs(i);break;}
        //for(int i=0;i<=25;i++) if(has[i]&&!vis[i]) {flag=0;break;}
        for(int i=0;i<=25;i++) if(has[i]&&find(i)!=root){flag=0;break;}
        if(flag){
            int more=0,less=0;
            for(int i=0;i<=25;i++){
                if(abs(outd[i]-ind[i])>=2) {flag=0;break;}
                if(outd[i]==ind[i]+1) more++;
                if(outd[i]==ind[i]-1) less++;
            }
            if(flag){
                if((more==0&&less==0)||(more==1&&less==1)) printf("Ordering is possible.
");
                else flag=0;
            }
        }
        if(flag==0) printf("The door cannot be opened.
");
    }
}