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
ENDOFINPUTSample Output
YES 1
NO
YES 10題目大意不在敖述，此題是一道典型的求無向圖中有無歐拉回路或歐拉通路的問題。首先是建圖：以房間為頂點，房間之間的門為邊建立無向圖。然後就是輸入問題，要求大家對字符串的有較好的處理能力，我用的是getchar()。然後請看代碼：
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<queue>
using namespace std ;
const int MAXN = 105 ;
const int INF = 0x7fffffff ;
int d[MAXN] ; // 建立頂點的度的數組
int main()
{
string s ;
int m , n ;
int sumt , sumj , sumd ;
while (cin >> s)
{
if(s == "START")
{
scanf("%d%d" , &m , &n) ;
getchar() ; // 處理剛才的回車，此處千萬不要忘記 ！！
memset(d , 0 , sizeof(d)) ;
int i ;
sumd = 0 ; // 統計邊的數目，即門的數目
int pan = 0 ; // 注意這個判斷變量的應用，請大家自己體會 ！！
for(i = 0 ; i < n ; i ++)
{
sumt = 0 ;
char t ;
while (1)
{
t = getchar() ;
if(t == '\n')
{
if(pan)
{
d[i] ++ ;
d[sumt] ++ ;
sumd ++ ;
pan = 0 ;
}
break ;
}
if(t == ' ')
{
d[i] ++ ;
d[sumt] ++ ;
sumd ++ ;
sumt = 0 ;
}
else
{
sumt = sumt * 10 + t - '0' ;
pan = 1 ;
}
}
}
}
if(s == "END")
{
int j ;
sumj = 0 ; // 統計奇度頂點的個數
for(j = 0 ; j < n ; j ++)
{
if(d[j] % 2 == 1)
{
sumj ++ ;
}
}
if(sumj > 2 || sumj == 1)
{
printf("NO\n") ;
}
else if(sumj == 0 && m != 0)
{
printf("NO\n") ;
}
else if(sumj == 2 && (d[m] % 2 != 1 || d[0] % 2 != 1))
{
printf("NO\n") ;
}
else if(sumj == 2 && d[m] % 2 == 1 && d[0] % 2 == 1 && m == 0)
{
printf("NO\n") ;
}
else
{
printf("YES %d\n" ,sumd) ;
}
}
if(s == "ENDOFINPUT")
{
break ;
}
}
return 0 ;
}
