无向图欧拉通路

poj1300

Door Man

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2020		Accepted: 801

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

题目大意：

一个房间有n个房间，编号0,1,2,3,……n-1;先输入m，n分别代表起点和房间数；

接下来n行，第i行给出一个数列，当某个数j大于i时表示房间i和j之间有个打开的门，走过之后就关上，且下次走的时候不能打开最后到达0房间；

首先判断是否联通

其次判断是否是欧拉回路还是欧拉通路

然后DFS找出结束的路径是否是0房间

程序：

#include"stdio.h"
#include"string.h"
#include"queue"
#include"stack"
#include"iostream"
#include"stdlib.h"
#define inf 99999999
#define M 111
using namespace std;
struct st
{
    int u,v,w,next,vis;
}edge[M*50];
int t,head[M],s[M],num;
void init()
{
    t=0;
    memset(head,-1,sizeof(head));
    memset(edge,0,sizeof(edge));
}
void add(int u,int v)
{
    edge[t].u=u;
    edge[t].v=v;
    edge[t].next=head[u];
    head[u]=t++;
}
void DFS(int u)
{
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(!edge[i].vis)
        {
            edge[i].vis=edge[i^1].vis=1;
            DFS(v);
        }
    }
    s[num++]=u;
}
int main()
{
    char ch[3333];
    while(scanf("%s",ch),strcmp(ch,"ENDOFINPUT")!=0)
    {
        int start,n,i,j;
        scanf("%d%d",&start,&n);
        getchar();
        init();
        for(i=0;i<n;i++)
        {
            gets(ch);
            int m=strlen(ch);
            int s=0;
            int flag=0;
            for(j=0;j<=m;j++)
            {
                if(ch[j]>='0'&&ch[j]<='9')
                {
                    flag=1;
                    s=s*10+ch[j]-'0';
                }
                else
                {
                    //printf("%d ",s);
                    if(s>i&&flag)
                    {
                        add(s,i);
                        add(i,s);
                        //printf("(%d %d)",i,s);
                    }

                    s=0;
                    flag=0;
                }
            }
            //printf("\n");
        }
        int cnt=0;
        for(i=0;i<n;i++)
        {
            int degree=0;
            for(j=head[i];j!=-1;j=edge[j].next)
            {
                degree++;
            }
            if(degree&1)
                cnt++;
        }
        while(scanf("%s",ch),strcmp(ch,"END")==0)
        {
            if(cnt!=0&&cnt!=2)
            {
                printf("NO\n");
                break;
            }
            num=0;
            DFS(start);
            /*for(i=num-1;i>=0;i--)
            {
                if(i==num-1)
                printf("%d",s[i]);
                else
                printf(" %d",s[i]);
            }
            printf("\n");*/
            if(s[0]==0)
            {
                printf("YES %d\n",num-1);
            }
            else
                printf("NO\n");
            break;
        }
    }
}