poj1300 Door Man（欧拉回路）

Door Man Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 2449   Accepted: 983 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:  Always shut open doors behind you immediately after passing through  Never open a closed door  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:  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).  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!  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 South Central USA 2002

很经典的欧拉回路问题，在图论算法书上看到的。。。

1.如果所有的房间都有偶数个门，那么就有欧拉回路。但是这种情况下必须从0出发才能够回到0；

2.如果有只有两个门是奇数，那么这两个门一定有一个是0，而且不从0出发。

3.其它情况都不行。

附上代码：

#include <stdio.h>
#include <string.h>
int main()
{
	char str[128];
	int door[25],m,n;
	while(gets(str))//gets可以接收空格
	{
		if(str[0]=='S')
		{
			sscanf(str,"%*s %d %d",&m,&n);//对于sscanf不懂的可以百度。这里*s是忽略str的第一个字符串（空格）
			for(int i=0;i<n;i++)
			door[i]=0;//计算每个房间有几道门
			int doors=0;//计算共有多少门
			for(int i=0;i<n;i++)
			{
				gets(str);
				int k=0,j;
				while(sscanf(str+k,"%d",&j)==1)//读取门
				{
					doors++;
					door[i]++;
					door[j]++;
					if(str[k]&&str[k]==' ') k++;//有空格+1就不说了吧
					while(str[k]&&str[k]!=' ') k++;//举个例子，如果房间号是12，就要+2了0.0
				}
					
			}
			gets(str);
			int odd=0,even=0;
			for(int i=0;i<n;i++)
			{
				if(door[i]%2)
				odd++;
				else
				even++;
			}
			if(odd==0&&m==0)
			printf("YES %d\n",doors);
			else if(odd==2&&door[m]%2&&door[0]%2&&m!=0)
			printf("YES %d\n",doors);
			else
			printf("NO\n",doors);
		}
		else
		break;
	}
	return 0;
}