zoj1395DoorMan(오라웨이 및 통로 존재 여부 판단)

Door Man
Time Limit: 1 Second     
Memory Limit: 32768 KB
Introduction
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 OutputYES 1
NO
YES 10
제목은 다음과 같다. 기호가 M인 방에서 출발하여 모든 문을 반복해서 지나가지 않고 닫은 후에 다시 열지 않고 (통과하지 않음), 마지막으로 자신의 방으로 돌아갈 수 있는가 하는 것이다.
두 가지 상황으로 나뉜다.
1:방마다 짝수 문이 있는데 이렇게 하면 반드시 오라회로가 있지만 반드시 M==0을 보증해야 한다. 왜냐하면 마지막에 0곳으로 돌아가야 하기 때문이다.
2: 방 2개에 홀수 문이 있고 다른 것은 짝수 문이 있으면 반드시 오라 통로가 있지만 M!=0, 그리고 0, M번 방에는 기점과 종점으로 홀수 문이 있다.
이 문제의 어려운 점은 데이터의 처리에 있다.입력에 빈칸이 있고 유효한 데이터로cin이 있을 수 없습니다. getchar () 가 하나씩 읽을 수 있을 뿐입니다. 그리고 한 줄에 문자도 있고 데이터도 있습니다. buf 그룹에 저장하고 sscanf로 읽으십시오...

#include<stdio.h>
#include<string.h>
int readLine(char* s)
{
    int L;
    for(L=0;(s[L]=getchar())!='
'&&s[L]!=EOF;L++);
    s[L]='\0';
    return L;
}
int main()
{
    int i,j;
    char buf[128];
    int M,N;
    int door[20];
    while(readLine(buf))
    {
        if(buf[0]=='S')
        {
            sscanf(buf,"%*s %d %d",&M,&N);
            for(i=0;i<N;i++)
              door[i]=0;
            int doors=0;
            for(i=0;i<N;i++)
            {
                readLine(buf);
                int k=0;
                while(sscanf(buf+k,"%d",&j)==1)
                {
                    door[i]++;
                    door[j]++;
                    doors++;
                    while(buf[k]&&buf[k]==' ') k++;
                    while(buf[k]&&buf[k]!=' ') k++;
                }
            }
            readLine(buf);
            int odd=0,even=0;
            for(i=0;i<N;i++)
            {
                if(door[i]%2==0) even++;
                else odd++;
            }
            if(odd==0&&M==0) printf("YES %d
",doors);
            else if(odd==2&&door[M]%2==1&&door[0]%2==1&&M!=0)
                 printf("YES %d
",doors);
            else printf("NO
");
        }
        else if(!strcmp(buf,"ENDOFINPUT"))
            break;
    }
    return 0;
}