Sightseeing tour--POJ 1637

1、题目类型:图论、混合图欧拉回路、Edmonds_Karp算法。

2、解题思路:(1)将图的无向边随便定向,计算每个点的入度和出度。如果有某个点出入度之差为奇数,那么肯定不存在欧拉回路;(2)构建流网络模型,Edmonds_Karp算法,察看是否有满流的分配。存在即有欧拉回路,没有就是没有欧拉回路。

3、注意事项:Edmonds_Karp算法中残留网络的更新。

4、参考博客:http://blog.csdn.net/bobten2008/archive/2009/11/11/4800084.aspx

5、实现方法:

  
    
#include < iostream >
#include
< cmath >
#include
< queue >
using namespace std;
#define Max 210
#define INF 99999999

int m,s,cnt;
int pre[Max],degree[Max];
int map[Max][Max],flag[Max][Max];
bool vis[Max];
queue
< int > Q;

int BFS()
{
int i,tmp,min,tmppre;
for (i = 0 ;i <= m + 1 ;i ++ )
{
vis[i]
= false ;
pre[i]
=- 1 ;
}
vis[
0 ] = true ;
while ( ! Q.empty())
Q.pop();
Q.push(
0 );
while ( ! Q.empty())
{
tmp
= Q.front();
Q.pop();
for (i = 1 ;i <= m + 1 ;i ++ )
{
if (i == tmp || vis[i] || ! map[tmp][i])
continue ;
vis[i]
= true ;
pre[i]
= tmp;
Q.push(i);
}
}
if (pre[m + 1 ] ==- 1 )
return - 1 ;
min
= INF,tmp = m + 1 ;
while ((tmppre = pre[tmp]) != - 1 )
{
if (map[tmppre][tmp] < min)
min
= map[tmppre][tmp];
tmp
= tmppre;
}
return min;
}

bool Edmonds_Karp( int cnt)
{
int i,tmp,val,tmppre,total;
while ((val = BFS()) !=- 1 )
{
tmp
= m + 1 ;
while ((tmppre = pre[tmp]) != - 1 )
{
map[tmppre][tmp]
-= val;
map[tmp][tmppre]
+= val;
flag[tmppre][tmp]
+= val;
flag[tmp][tmppre]
=- flag[tmppre][tmp];
tmp
= tmppre;
}
}
total
= 0 ;
for (i = 1 ;i <= m;i ++ )
total
+= flag[ 0 ][i];
return total == cnt;
}
int main()
{
int i,n,start,end,type;
bool mark;
cin
>> n;
while (n -- )
{
cin
>> m >> s;
mark
= true ;
memset(degree,
0 , sizeof (degree));
memset(map,
0 , sizeof (map));
memset(flag,
0 , sizeof (flag));
for (i = 0 ;i < s;i ++ )
{
cin
>> start >> end >> type;
degree[start]
++ ;
degree[end]
-- ;
if (type != 1 )
map[start][end]
++ ;
}
for (i = 1 ;i <= m;i ++ )
{
if (abs(degree[i]) % 2 == 1 )
{
mark
= false ;
break ;
}
else
degree[i]
= degree[i] / 2 ;
}
if (mark)
{
cnt
= 0 ;
for (i = 1 ;i <= m;i ++ )
{
if (degree[i] < 0 )
map[i][m
+ 1 ] = abs(degree[i]);
else if (degree[i] > 0 )
{
map[
0 ][i] = degree[i];
cnt
+= degree[i];
}
}
if ( ! Edmonds_Karp(cnt))
mark
= false ;
}
if (mark)
cout
<< " possible " << endl;
else
cout
<< " impossible " << endl;
}
return 1 ;
}

 

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