POJ 1300 Door Man (有向图欧拉通路)

Door Man

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2330		Accepted: 923

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. 
   
2. Always shut open doors behind you immediately after passing through
   
3. Never open a closed door
   
4. 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. 
   
2. 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).
   
3. 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!
   
4. 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

题目链接：http://poj.org/problem?id=1300

题目大意：妈蛋，题意绕死人，简单地说就是一个人从第一个房间开始每过一个房间就把那个房间的门关上，关上的门不得打开，问最后能不能回到自己的房间(0号)，如果可以输出YES和关闭的门的数量，start下的第i行表示从第m+i个房间能到达的其他房间编号
并且已知任意两个门可连通

题目分析：最简单的了。。。只是题意复杂，告诉连通，既不用判连通，又不要输出路径，没什么好说的，这题就是考你英语和读入数据的能力

#include <cstdio>
#include <cstring>

int get(char *s)
{
    int len;
    for(len = 0; (s[len] = getchar()) != '\n' && s[len] != EOF; len++);
    s[len] = 0;
    return len;
}

int main()
{
    char s[150], t[150];
    int st, n, door[25];
    while(get(s))
    {
        if(s[0] == 'S')
        {
            sscanf(s, "%s %d %d", t, &st, &n);
            memset(door, 0, sizeof(door));
            int sum = 0, j;
            for(int i = 0; i < n; i++)
            {
                get(s);
                int cnt = 0;
                while(sscanf(s + cnt, "%d", &j) == 1)
                {
                    sum++;
                    door[i] ++;
                    door[j] ++;
                    while(s[cnt] && s[cnt] == ' ')
                        cnt++;
                    while(s[cnt] && s[cnt] != ' ')
                        cnt++;
                }
            }
            get(s);
            int odd = 0, even = 0;
            for(int i = 0; i < n; i++)
            {
                if(door[i] % 2)
                    odd++;
                else
                    even++;
            }
            if(odd == 0 && st == 0)
                printf("YES %d\n", sum);
            else if(odd == 2 && door[st] % 2 == 1 && door[0] % 2 == 1 && st != 0)
                printf("YES %d\n", sum);
            else
                printf("NO\n");
        }
        else if(!strcmp(s, "ENDOFINPUT"))
            break;
    }
}