zoj1395DoorMan (判断是否存在欧拉回路和通路)

Door Man

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Time Limit: 1 Second      Memory Limit: 32768 KB

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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 Output

YES 1

NO

YES 10

题意是这样的：从标号为M的房间出发，可否不重复的经过每一扇门，将其关闭后，不再打开（不再通过），最后回到自己的房间（0号）。

分两种情况：

1：每个房间都有偶数扇门，这样必有欧拉回路，但是必须保证M==0 ，因为最后要回到0处；

2：有2个房间有奇数扇门，其他都有偶数扇门，则必有欧拉通路，但M!=0，且0，M号房间有奇数扇门，作为起点和终点；

本题困难处在于数据的处理。输入中有空格，且作为有效数据不能有cin，只能是有getchar( )一个个读入，且一行中既有字符，又有数据，先保存到buf数组中，再用sscanf读入……

#include<stdio.h>#include<string.h>int readLine(char* s){    int L;    for(L=0;(s[L]=getchar())!='\n'&&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\n",doors);            else if(odd==2&&door[M]%2==1&&door[0]%2==1&&M!=0)                 printf("YES %d\n",doors);            else printf("NO\n");        }        else if(!strcmp(buf,"ENDOFINPUT"))            break;    }    return 0;}