【Codeforces Beta Round #36】Codeforces 36E Two Paths

Once archaeologists found m mysterious papers, each of which had a

pair of integers written on them. Ancient people were known to like

writing down the indexes of the roads they walked along, as «a b» or

«b a», where a, b are the indexes of two different cities joint by the

road . It is also known that the mysterious papers are pages of two

travel journals (those days a new journal was written for every new

journey).

During one journey the traveler could walk along one and the same road

several times in one or several directions but in that case he wrote a

new entry for each time in his journal. Besides, the archaeologists

think that the direction the traveler took on a road had no effect

upon the entry: the entry that looks like «a b» could refer to the

road from a to b as well as to the road from b to a.

The archaeologists want to put the pages in the right order and

reconstruct the two travel paths but unfortunately, they are bad at

programming. That’s where you come in. Go help them!

Input The first input line contains integer m (1 ≤ m ≤ 10000). Each of

the following m lines describes one paper. Each description consists

of two integers a, b (1 ≤ a, b ≤ 10000, a ≠ b).

Output In the first line output the number L1. That is the length of

the first path, i.e. the amount of papers in its description. In the

following line output L1 space-separated numbers — the indexes of the

papers that describe the first path. In the third and fourth lines

output similarly the length of the second path L2 and the path itself.

Both paths must contain at least one road, i.e. condition L1 > 0 and

L2 > 0 must be met. The papers are numbered from 1 to m according to

the order of their appearance in the input file. The numbers should be

output in the order in which the traveler passed the corresponding

roads. If the answer is not unique, output any.

If it’s impossible to find such two paths, output «-1».

Don’t forget that each paper should be used exactly once, i.e

L1 + L2 = m.

分类讨论比较麻烦。

如果有多于两个联通块或者只有一条边，一定无解。

如果有两个联通块，那么分别求欧拉路。

如果有一个联通块，如果本身就有欧拉路直接求出然后随意断开。此外如果入度为奇数的点【以下简称奇点】有4个，还可以两笔画。做法是连接两个奇点，求剩下的两个奇点的欧拉路，再从添加的这条边断开。

求欧拉路可以用Fleury算法dfs一遍解决。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
vector<pair<int,int> > to[10010];
int fa[10010],bel[10010],size[10010],sta[10010],top,m,n,clo;
bool vis[10010];
int find(int x)
{
return x==fa[x]?x:fa[x]=find(fa[x]);
}
bool init()
{
int i,x,y;
scanf("%d",&m);
for (i=1;i<=10000;i++)
fa[i]=i;
for (i=1;i<=m;i++)
{
scanf("%d%d",&x,&y);
n=max(n,max(x,y));
fa[find(x)]=find(y);
to[x].push_back(make_pair(y,i));
to[y].push_back(make_pair(x,i));
}
for (i=1;i<=n;i++)
if (to[i].size()&&fa[i]==i)
bel[i]=++clo;
for (i=1;i<=n;i++)
if (to[i].size())
bel[i]=bel[find(i)];
for (i=1;i<=n;i++)
if (to[i].size())
size[bel[i]]+=to[i].size();
for (i=1;i<=clo;i++)
size[i]/=2;
}
void dfs(int u)
{
int i,v,x;
for (i=0;i<to[u].size();i++)
if (!vis[x=to[u][i].second])
{
v=to[u][i].first;
vis[x]=1;
dfs(v);
sta[++top]=x;
}
}
void out1()
{
int i;
printf("1\n%d\n%d\n",sta[1],m-1);
for (i=2;i<=m;i++)
printf("%d%c",sta[i],i==m?'\n':' ');
}
void out2()
{
int i,p;
for (p=1;;p++)
if (sta[p]==m+1)
break;
printf("%d\n",p-1);
for (i=1;i<p;i++)
printf("%d%c",sta[i],i==p-1?'\n':' ');
printf("%d\n",m-p+1);
for (i=p+1;i<=m+1;i++)
printf("%d%c",sta[i],i==m+1?'\n':' ');
}
void solve1()
{
int i,x,y,z,cnt=0;
for (i=1;i<=n;i++)
if (to[i].size()&1)
cnt++;
switch (cnt)
{
case 0:
for (i=1;i<=n;i++)
if (to[i].size())
{
dfs(i);
break;
}
break;
case 2:
for (i=1;i<=n;i++)
if (to[i].size()&1)
{
dfs(i);
break;
}
break;
case 4:
{
x=y=-1;
for (i=1;i<=n;i++)
if (to[i].size()&1)
{
if (x==-1) x=i;
else
{
y=i;
break;
}
}
to[x].push_back(make_pair(y,m+1));
to[y].push_back(make_pair(x,m+1));
size[bel[x]]++;
for (i=1;i<=n;i++)
if (to[i].size()&1)
{
dfs(i);
break;
}
break;
}
default:
printf("-1\n");
return;
break;
}
if (cnt==4) out2();
else out1();
}
void out0()
{
int i;
printf("%d\n",top);
for (i=1;i<=top;i++)
printf("%d%c",sta[i],i==top?'\n':' ');
top=0;
}
void solve2()
{
int i,cnt1=0,cnt2=0;
for (i=1;i<=n;i++)
if (to[i].size()&1)
{
if (bel[i]==1) cnt1++;
else cnt2++;
}
if (cnt1>2||cnt2>2)
{
printf("-1\n");
return;
}
if (!cnt1)
{
for (i=1;i<=n;i++)
if (to[i].size()&&bel[i]==1)
{
dfs(i);
break;
}
}
else
{
for (i=1;i<=n;i++)
if (to[i].size()&1&&bel[i]==1)
{
dfs(i);
break;
}
}
out0();
if (!cnt2)
{
for (i=1;i<=n;i++)
if (to[i].size()&&bel[i]==2)
{
dfs(i);
break;
}
}
else
{
for (i=1;i<=n;i++)
if (to[i].size()&1&&bel[i]==2)
{
dfs(i);
break;
}
}
out0();
}
int main()
{
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
init();
if (m==1)
{
printf("-1\n");
return 0;
}
switch (clo)
{
case 1:solve1();break;
case 2:solve2();break;
default: printf("-1\n");break;
}
}