Play on Words UVA - 10129 欧拉路径

关于欧拉回路和欧拉路径

定义：
欧拉回路：每条边恰好只走一次，并能回到出发点的路径
欧拉路径：经过每一条边一次，但是不要求回到起始点

①首先看欧拉回路存在性的判定：

一、无向图
每个顶点的度数都是偶数，则存在欧拉回路。

二、有向图（所有边都是单向的）
每个节顶点的入度都等于出度，则存在欧拉回路。

②.欧拉路径存在性的判定

一。无向图
一个无向图存在欧拉路径，当且仅当   该图所有顶点的度数为偶数   或者  除了两个度数为奇数外其余的全是偶数。

二。有向图
一个有向图存在欧拉路径，当且仅当  该图所有顶点的入度等于出度 或者 一个点出度比入度多一（起点） 一个点入度比出度多一（终点） 其他都为入度等于出度。

在已经知道存在的情况下，下面的程序用于输出路径；

如果是输出欧拉道路，那么参数必须是道路的起点； 另外打印的顺序是逆序的，因此真正使用时候，用push存入(u,v)；

void euler（int u){

for(Int v = 0; v < n; v++){

if(G[u][v]&&!vis[u][v]){

vis[u][v] = vis[v][u]=1;
          euler(v);
          printf("%d %d\n",u,v);

}

}

}

上面的代码只适用于无向图，改成有向图只需要把vis[u][v] =vis[v][u] = 1 改成 vis[u][v]即可；

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 `

m

otorola

'.

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 rst line of the input le.

Each test case begins with a line containing a single integer number

N

that indicates the number of

plates (1

N

100000). Then exactly

N

lines 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 rst 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.

 

 

/**
题目：Play on Words UVA - 10129
链接：https://vjudge.net/problem/UVA-10129
题意：给定n个单词，排成一行，使单词的首部字母和上一个单词（如果存在）的尾部字母相同。问是否存在这样的一行序列满足要求。
 
思路：
首先保证不考虑方向的情况下图是连通的。
 
判断有向图是否存在欧拉路径。
1，所有的点入度等于出度。
2，一个点入度比出度多一，一个点出度比入度多一，其他点入度等于出度。
 
把每一个word的首字母和尾字母连一条有向的边。
*/
 
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long LL;
const int mod=1e9+;
const int maxn=1e2+;
const double eps = 1e-;
int T, n;
char s[];
int in[], ot[];
int st[];
int vis[];
int Find(int x)
{
    if(x==st[x]) return x;
    return st[x] = Find(st[x]);
}
void Merge(int x,int y)
{
    int fx = Find(x);
    int fy = Find(y);
    if(fx>fy){
        st[fx] = fy;
    }else st[fy] = fx;
}
int main()
{
    cin>>T;
    while(T--)
    {
        scanf("%d",&n);
        memset(in, , sizeof in);
        memset(ot, , sizeof ot);
        memset(vis, , sizeof vis);
        for(int i = ; i < ; i++) st[i] = i;
        for(int i = ; i < n; i++){
            scanf("%s",s);
            int len = strlen(s);
            vis[s[]-'a'] = vis[s[len-]-'a'] = ;
            Merge(s[]-'a',s[len-]-'a');
            ot[s[]-'a']++;
            in[s[len-]-'a']++;
        }
 
        int r;
        for(int i = ; i < ; i++){
            if(vis[i]){
                r = Find(i);
            }
        }
        int sign = ;
        for(int i = ; i < ; i++){
            if(vis[i]){
                if(Find(i)!=r){
                    sign = ; break;
                }
            }
        }
        if(sign){///不考虑方向，没有连通。
            printf("The door cannot be opened.\n"); continue;
        }
 
        int flag1, flag2, flag3;
        flag1 = flag2 = flag3 = ;///分别表示入度等于出度的点数，入度比出度多一的点数，出度比入度多一的点数。
        for(int i = ; i < ; i++){
            if(in[i]==ot[i]){
                flag1++;
            }
            if(in[i]-ot[i]==){
                flag2++;
            }
            if(ot[i]-in[i]==){
                flag3++;
            }
        }
        if(flag1==||((flag1==)&&(flag2==)&&(flag3==))){
            printf("Ordering is possible.\n");
        }else
         printf("The door cannot be opened.\n");
    }
    return ;
}