POJ1300 Door Man —— 欧拉回路（无向图）

题目链接：http://poj.org/problem?id=1300

Door Man

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2899		Accepted: 1176

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

题解：

典型的欧拉回路判定，输出YES的情况有两种（已知起点为m， 终点为0）：

1.所有点的度数都为偶数，并且起点终点都为0。

2.有两个点的度数为奇数，其他都为偶数，并且这两个点为起点和终点。

欧拉回路：

首要条件：连通

无向图：
情况1：每个结点均为偶度结点。
情况2：有两个结点为奇度结点，其他结点均为偶度结点。

有向图：
情况1：每个结点的入度等于出度。
情况2：有一个结点入度等于出度+1，有一个结点出度等于入度+1，其他结点的入度等于出度。

代码如下：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long LL;
const double EPS = 1e-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+7;
const int MAXN = 20+10;

int n, m, deg[MAXN];
char str[100];

int main()
{
    while(scanf("%s", str) && strcmp(str, "ENDOFINPUT"))
    {
        scanf("%d%d",&m, &n);
        memset(deg, 0, sizeof(deg));
        int ans = 0;
        getchar();
        for(int i = 0; i<n; i++)
        {
            gets(str);
            for(int j = 0; j<strlen(str); j++)
            {
                if(str[j]==' ') continue;
                ans++;
                deg[i]++;
                deg[str[j]-'0']++;
            }
        }
        gets(str);

        int cnt = 0;
        for(int i = 0; i<n; i++)
            if(deg[i]&1) cnt++;

        if(cnt==0 && m==0)  //没有奇点，则起点和终点都必须为0
            printf("YES %d
", ans);
        else if(cnt==2 && m!=0 &&(deg[m]&1)&&(deg[0]&1) )  //有两个奇点， 则这两个奇点必须为起点和终点
            printf("YES %d
", ans);
        else    //奇点个数等于1或大于2，则不存在欧拉回路
            printf("NO
");
    }
}