POJ1300（欧拉回路）

Door Man

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2139		Accepted: 858

Description

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

Source

field=source&key=South+Central+USA+2002" style="text-decoration:none">South Central USA 2002

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题目描写叙述:

你是一座大庄园的管家。

庄园有非常多房间,编号为 0、1、2、3,...。

你的主人是一个心不在焉的人,常常沿着走廊任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。你的问题是是否能找到一条路径经过全部开着门的房间,并使得:

这题字符处理挺麻烦的。

。

。。

1) 通过门后马上把门关上;

2) 关上了的门不再打开;

3) 最后回到你自己的房间(房间 0)，而且全部的门都已经关闭了。

以房间为顶点、连接房间之间的门为边构造图。依据题目的意思,输入文件里每一个測试数据所构造的图都是连通的。本题实际上是推断一个图中是否存在欧拉回路或欧拉通路,要分两种情况考虑:

1:

假设全部的房间都有偶数个门(通往其它房间),那么有欧拉回路,能够从 0 号房间出发,回到 0 号房间。可是这样的情况下,出发的房间必须为 0,由于要求回到 0 号房间。

2:

有两个房间的门数为奇数,其余的都是偶数,假设出发的房间和 0 号房间的门数都是奇数,那么也能够从出发的房间到达 0 号房间,而且满足题目要求。可是不能从房间 0 出发,必须从还有一个门数为奇数的房间出发。

#include <cstdio>
#include <cstring>
int readLine( char* s )
{
	int L;
	for( L=0; ( s[L]=getchar() ) != '\n' && s[L] != EOF; L++ );
		s[L] = 0;
	return L;
}
int main( )
{
	int i, j;
	char buf[128];
	int M, N;
	int door[20];
	while( readLine(buf) )
	{
		if( buf[0]=='S' )
		{
			sscanf( buf, "%*s %d %d", &M, &N );
			for( i=0; i < N; i++ )
			door[i] = 0;
			int doors = 0;
			for( i=0; i<N; i++ )
			{
				readLine(buf);
				int k = 0; 

				while( sscanf(buf + k, "%d", &j) == 1 )
				{
					doors++;
					door[i]++;
					door[j]++;
					while( buf[k] && buf[k] == ' ' ) k++;
					while( buf[k] && buf[k] != ' ' ) k++;
				}
			}
			readLine( buf );

			int odd = 0, even = 0;
			for( i=0; i<N; i++ )
			{
				if( door[i]%2==0 ) even++;
				else odd++;
			}
			if( odd==0 && M==0 ) printf( "YES %d\n", doors );
			else if( odd==2 && door[M]%2==1 && door[0]%2==1 && M!=0 )
				printf( "YES %d\n", doors );
			else printf( "NO\n" );
		}
		else if( !strcmp(buf, "ENDOFINPUT") )
			break;
	}
	return 0;
}