USACO-Section 3.3-PROB Riding The Fences

Riding the Fences

Farmer John owns a large number of fences that must be repaired annually. He traverses the fences by riding a horse along each and every one of them (and nowhere else) and fixing the broken parts.

Farmer John is as lazy as the next farmer and hates to ride the same fence twice. Your program must read in a description of a network of fences and tell Farmer John a path to traverse each fence length exactly once, if possible. Farmer J can, if he wishes,
start and finish at any fence intersection.

Every fence connects two fence intersections, which are numbered inclusively from 1 through 500 (though some farms have far fewer than 500 intersections). Any number of fences (>=1) can meet at a fence intersection. It is always possible to ride from any
fence to any other fence (i.e., all fences are "connected").

Your program must output the path of intersections that, if interpreted as a base 500 number, would have the smallest magnitude.

There will always be at least one solution for each set of input data supplied to your program for testing.

PROGRAM NAME: fence

INPUT FORMAT

Line 1:	The number of fences, F (1 <= F <= 1024)
Line 2..F+1:	A pair of integers (1 <= i,j <= 500) that tell which pair of intersections this fence connects.

SAMPLE INPUT (file fence.in)

9
1 2
2 3
3 4
4 2
4 5
2 5
5 6
5 7
4 6

OUTPUT FORMAT

The output consists of F+1 lines, each containing a single integer. Print the number of the starting intersection on the first line, the next intersection's number on the next line, and so on, until the final intersection on the last line. There might be
many possible answers to any given input set, but only one is ordered correctly.

SAMPLE OUTPUT (file fence.out)

1
2
3
4
2
5
4
6
5
7

欧拉回路

深搜，对于当前的点，把所有点从小到大搜索，找到和它相连的，找到一个之后删除它们之间的连线，并去搜索新的那个点，如果没有找到点和它相连，那么就把这个点加入输出队列。
输出时反着输出即可。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#define name "fence"
using namespace std;
int edge[505][505],du[255];
bool have[505];
int m,s,maxn,minn=1e9;
int ans[255],anst;
void dfs(int u)
{
for (int v=minn;v<=maxn;v++)
{
if (edge[u][v]!=0)
{
edge[u][v]--;
edge[v][u]--;
dfs(v);
}
}
ans[++anst]=u;
}
int main()
{
freopen(name ".in","r",stdin);
freopen(name ".out","w",stdout);
int i,j;
cin>>m;	int u,v;
for (i=1;i<=m;i++)
{
scanf("%d%d",&u,&v);
du[u]++;du[v]++;
have[u]=1;have[v]=1;
maxn=max(maxn,u);maxn=max(maxn,v);
minn=min(minn,u);minn=min(minn,v);
edge[u][v]++;edge[v][u]++;
}
for (i=1;i<=maxn;i++)
if (have[i]&&du[i]%2==1)
{s=i;break;}
if (s==0) s=minn;
dfs(s);
for (i=anst;i>=1;i--)
cout<<ans[i]<<endl;
return 0;
}