C - Door Man（欧拉回路_格式控制）

现在你是一个豪宅的管家，因为你有个粗心的主人，所以需要你来帮忙管理，输入会告诉你现在一共有多少个房间，然后会告诉你从哪个房间出发，
你的任务就是从出发的房间通过各个房间之间的通道，来把所有的门都关上，然后最后回到0的房间。

You are a butler in a large mansion. This mansion has so many rooms that they are merely referred to by number (room 0, 1, 2, 3, etc...). Your master is a particularly absent-minded lout and continually leaves doors open throughout a particular floor of the house. Over the years, you have mastered the art of traveling in a single path through the sloppy rooms and closing the doors behind you. Your biggest problem is determining whether it is possible to find a path through the sloppy rooms where you:

1. Always shut open doors behind you immediately after passing through
2. Never open a closed door
3. End up in your chambers (room 0) with all doors closed

In this problem, you are given a list of rooms and open doors between them (along with a starting room). It is not needed to determine a route, only if one is possible.

Input

Input to this problem will consist of a (non-empty) series of up to 100 data sets. Each data set will be formatted according to the following description, and there will be no blank lines separating data sets.
A single data set has 3 components:

1. Start line - A single line, "START M N", where M indicates the butler's starting room, and N indicates the number of rooms in the house (1 <= N <= 20).
2. Room list - A series of N lines. Each line lists, for a single room, every open door that leads to a room of higher number. For example, if room 3 had open doors to rooms 1, 5, and 7, the line for room 3 would read "5 7". The first line in the list represents room 0. The second line represents room 1, and so on until the last line, which represents room (N - 1). It is possible for lines to be empty (in particular, the last line will always be empty since it is the highest numbered room). On each line, the adjacent rooms are always listed in ascending order. It is possible for rooms to be connected by multiple doors!
3. End line - A single line, "END"

Following the final data set will be a single line, "ENDOFINPUT".

Note that there will be no more than 100 doors in any single data set.

Output

For each data set, there will be exactly one line of output. If it is possible for the butler (by following the rules in the introduction) to walk into his chambers and close the final open door behind him, print a line "YES X", where X is the number of doors he closed. Otherwise, print "NO".

Sample Input

START 1 2
1

END
START 0 5
1 2 2 3 3 4 4

END
START 0 10
1 9
2
3
4
5
6
7
8
9

END
ENDOFINPUT

Sample Output

YES 1
NO
YES 10

题目描述：

你是一座大庄园的管家。庄园有很多房间，编号为0、1、2、3，...。你的主人是一个心不在焉的人，经常沿着走廊随意地把房间的门打开。多年来，你掌握了一个诀窍：沿着一个通道，穿过这些大房间，并把房门关上。你的问题是能否找到一条路径经过所有开着门的房间，并使得：
1) 通过门后立即把门关上；
2) 关上了的门不再打开；
3) 最后回到你自己的房间（房间0），并且所有的门都已经关闭了。
在本题中，给定房间列表，及连通房间的、开着的门，并给定一个起始房间，判断是否存在这样的一条路径。不需要输出这样的路径，只需判断是否存在。假定任意两个房间之间都是连通的（可能需要经过其他房间）。
输入描述：
输入文件包含多个（最多可达100 个）测试数据，每个测试数据之间没有空行隔开。
每个测试数据包括3 部分：
起始行－格式为“START M N”，其中M 为管理员起始所处的房间号，N 为房间的总数（1≤N≤20）；
房间列表－一共N 行，每行列出了一个房间通向其他房间的房间号（只需列出比它的号码大的房间号，可能有多个，按升序排列），比如房间3 有门通向房间1、5、7，则房间3 的信息行内容为“5 7”，第一行代表房间0，最后一行代表行间N-1。有可能有些行为空行，当然最后一行肯定是空行，因为N-1 是最大的房间号；两个房间之间可能有多扇门连通。
终止行－内容为"END"。
输入文件最后一行是"ENDOFINPUT"，表示输入结束。
输出描述：

每个测试数据对应一行输出，如果能找到一条路关闭所有的门，并且回到房间0，则输出"YESX"，X 是他关闭的门的总数，否则输出"NO"。

欧拉回路小题

• 1) 如果所有的房间都有偶数个门（通往其他房间），那么有欧拉回路，可以从0 号房间出发，回到0 号房间。但是这种情况下，出发的房间必须为0，因为要求回到0 号房间。
• 2) 有两个房间的门数为奇数，其余的都是偶数，如果出发的房间和0 号房间的门数都是奇数，那么也可以从出发的房间到达0 号房间，并且满足题目要求。但是不能从房间0 出发，必须从另一个门数为奇数的房间出发。

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <algorithm>
 4 using namespace std;
 5 int main()
 6 {
 7     int mp[200];
 8     char s[30];
 9     int n,m;
10     while(scanf("%s",s),strcmp(s,"ENDOFINPUT"))
11     {
12         memset(mp,0,sizeof(mp));
13         scanf("%d%d",&n,&m);
14         char ch=getchar();
15         for(int i=0;i<m;i++)
16         {
17             ch=getchar();
18             while(ch!='\n')
19             {
20                 if(ch!=' ')
21                 {
22                     mp[i]+=1;
23                     mp[ch-'0']++;
24                 }
25                 ch=getchar();
26             }
27         }
28         int flag=0,sum=0;
29         for(int i=0;i<m;i++)
30         {
31             if(mp[i]%2)
32             flag++;
33             sum+=mp[i];
34         }
35         /*
36          欧拉回路的特点是入度等于出度，
37          若起始点为0，则需要所有入度都等于出度，
38          若起始点不是0，
39          则需要非起始点和0点以外都有偶数个出入边。
40          起始点和0点都有奇数个出入边。
41         */
42         if((flag==0&&n==0)||(flag==2&&(mp[0]%2)&&n))
43         printf("YES %d\n",sum/2);
44         else
45             printf("NO\n");
46         scanf("%s",s);
47     }
48 }