欧拉回路-Door Man 分类： 图论 POJ 2015-08-06 10:07 4人阅读 评论(0) 收藏

Door Man

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 2476 Accepted: 1001

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

题意:给你起点m和房间的个数n,接下来有n行,第一行代表房间0与之相连的房间编号(只列出比它大的房间编号,按升序),一个房间可能有多个门,问可从起点出发将所有的门关上,并回到房间0,如果可以关上输出”YES X”,X代表关上的门的数量,不能输出”NO”.

方法:一道无向图的欧拉回路的题,判断是不是欧拉回路,就看图的度,如果度为奇数的个数为零,则无论从哪一个点出发都可以回到0,如果度为奇数的为2个,则这两个点一个为起点一个为终点,需要判断起点是不是度为奇数并且0的度也为零,(m!=0).其余的情况都不可以.这个题的数据处理比较麻烦点.

#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-9
#define LL long long
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CRR fclose(stdin)
#define CWW fclose(stdout)
#define RR freopen("input.txt","r",stdin)
#pragma comment(linker, "/STACK:102400000")
#define WW freopen("output.txt","w",stdout)

const int MAX = 11000;

char buf[MAX];
char s[50];
int Du[120];

int ReadLine()
{
    int L;
    for(L=0;(buf[L]=getchar())!='\n'&&buf[L]!=EOF;L++);
    buf[L]='\0';
    return L;
}
int  main()
{
    int n,m;
    int door;
    while(ReadLine())
    {
        if(buf[0]=='S')
        {
            sscanf(buf,"%s %d %d",s,&m,&n);
            memset(Du,0,sizeof(Du));
            door=0;
            for(int i=0;i<n;i++)
            {
                ReadLine();
                int k=0;
                int v;
                while(sscanf(buf+k,"%d",&v)==1)
                {
                    door++;
                    Du[i]++;
                    Du[v]++;
                    while(buf[k]&&buf[k++]==' ');
                    while(buf[k]&&buf[k++]!=' ');
                }
            }
            ReadLine();
            int ood=0;
            for(int i=0;i<n;i++)
            {
                if(Du[i]&1)
                {
                    ood++;
                }
            }
            if((ood==0&&m==0)||(ood==2&&Du[m]&1&&Du[0]&1&&m))
            {
                printf("YES %d\n",door);
            }
            else
            {
                printf("NO\n");
            }
        }
        else if(!strcmp(buf,"ENDOFINPUT"))
        {
            break;
        }
    }
    return 0;
}

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