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： 1) 通过门后马上把门关上。 2) 关上了的门不再打开。 3) 后回到你自己的房间（房间 0），而且全部的门都已经关闭了。 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。推断是否存在
这种一条路径。不须要输出这种路径。仅仅需推断是否存在。
假定随意两个房间之间都是连通 的（可能须要经过其它房间）。
输入描写叙述：
输入文件包括多个（多可达 100 个）測试数据，每一个測试数据之间没有空行隔开。
每一个測试数据包括 3部分： 起始行－格式为“START M N”，当中 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它房间的房间号（仅仅需列出比它的号码大 的房间号，可能有多个，按升序排列），比方房间 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，后一行代表行间 N-1。
有可能有些行为空行。当然后一行肯 定是空行，由于 N-1 是大的房间号；两个房间之间可能有多扇门连通。
终止行－内容为"END"。
输入文件后一行是"ENDOFINPUT"，表示输入结束。
输出描写叙述：
每一个測试数据相应一行输出，假设能找到一条路关闭全部的门，而且回到房间 0，则输出"YES X"。X是他关闭的门的总数。否则输出"NO"。
就是推断是否能构成欧拉通路·咯
无向图存在欧拉通路的充要条件：
1.
是连通图
2.
奇度节点个数为0或2,当中为0时为欧拉回路,为2时是以这两个点节点为端点的欧拉通路
#include<cstdio>
#include<cstring>
#include<cctype>
const int N = 21;
using namespace std;
int main()
{
int door, cnt, m, n, deg[N];
char s[N], c;
while(~scanf("%s", s), strcmp(s, "ENDOFINPUT"))
{
memset(deg, 0, sizeof(deg));
door = cnt = 0;
scanf("%d %d\n", &m, &n);
for(int i = 0; i < n; ++i)
{
while(scanf("%c", &c))
{
if(isdigit(c)) ++door, ++deg[i], ++deg[c - ‘0‘];
else if(c == ‘ ‘) continue;
else break;
}
if(deg[i] % 2) ++cnt;
}
scanf("%s", s);
bool ok = (cnt == 0 && m == 0 ) || (m && cnt == 2 && deg[m] % 2 );
if(ok) printf("YES %d\n", door);
else printf("NO\n");
}
return 0;
}
Door Man
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