Sightseeing tour--POJ 1637

最新推荐文章于 2019-06-11 11:44:00 发布

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最新推荐文章于 2019-06-11 11:44:00 发布

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原文链接：http://www.cnblogs.com/yongze103/archive/2010/08/12/1798410.html

本文介绍了一种利用图论解决混合图欧拉回路问题的方法，并详细阐述了如何通过Edmonds-Karp算法来判断是否存在满流分配，进而确定欧拉回路的存在性。文章还提供了具体的C++实现代码。

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.youkuaiyun.com/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 ;
}

转载于:https://www.cnblogs.com/yongze103/archive/2010/08/12/1798410.html