标签：欧拉回路
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Door Man
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:
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:
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 |
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题目描述:
你是一座大庄园的管家。庄园有很多房间,编号为 0、1、2、3,...。你的主人是一个心不在焉的人,经常沿着走廊随意地把房间的门打开。多年来,你掌握了一个诀窍:沿着一个通道,穿过这些大房间,并把房门关上。你的问题是能否找到一条路径经过所有开着门的房间,并使得:
这题字符处理挺麻烦的。。。。
1) 通过门后立即把门关上;
2) 关上了的门不再打开;
3) 最后回到你自己的房间(房间 0)，并且所有的门都已经关闭了。
以房间为顶点、连接房间之间的门为边构造图。根据题目的意思,输入文件中每个测试数据所构造的图都是连通的。本题实际上是判断一个图中是否存在欧拉回路或欧拉通路,要分两种情况考虑:
1:
如果所有的房间都有偶数个门(通往其他房间),那么有欧拉回路,可以从 0 号房间出发,回到 0 号房间。但是这种情况下,出发的房间必须为 0,因为要求回到 0 号房间。
2:
有两个房间的门数为奇数,其余的都是偶数,如果出发的房间和 0 号房间的门数都是奇数,那么也可以从出发的房间到达 0 号房间,并且满足题目要求。但是不能从房间 0 出发,必须从另一个门数为奇数的房间出发。
#include <cstdio>
#include <cstring>
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 )
{
doors++;
door[i]++;
door[j]++;
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;
}
标签：欧拉回路
原文地址：http://blog.csdn.net/acvcla/article/details/40357561
