Farmer John owns a large number of fences that must be repairedannually. He traverses the fences by riding a horse along each andevery 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 samefence twice. Your program must read in a description of a network offences and tell Farmer John a path to traverse each fence length exactlyonce, if possible. Farmer J can, if he wishes,
start and finish at anyfence intersection.

Every fence connects two fence intersections, which are numberedinclusively from 1 through 500 (though some farms have far fewer than500 intersections). Any number of fences (>=1) can meet at a fenceintersection. It is always possible to ride from any fence
to any otherfence (i.e., all fences are "connected").

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

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

PROGRAM NAME: fence

INPUT FORMAT

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, thenext intersection's number on the next line, and so on, until the finalintersection on the last line. There might be many
possible answers toany given input set, but only one is ordered correctly.

SAMPLE OUTPUT (file fence.out)

1
2
3
4
2
5
4
6
5
7

解决方案

找出起始点，然后用从起始点开始用深搜寻找欧拉路径，唯一需要注意的是两个点之间可能有多条边连接。

运行结果

Executing...
   Test 1: TEST OK [0.003 secs, 4480 KB]
   Test 2: TEST OK [0.003 secs, 4480 KB]
   Test 3: TEST OK [0.005 secs, 4480 KB]
   Test 4: TEST OK [0.003 secs, 4480 KB]
   Test 5: TEST OK [0.005 secs, 4480 KB]
   Test 6: TEST OK [0.008 secs, 4480 KB]
   Test 7: TEST OK [0.011 secs, 4480 KB]
   Test 8: TEST OK [0.014 secs, 4480 KB]

All tests OK.

通过代码

/*
ID: c1033311
LANG: C++
TASK: fence
*/

#include<stdio.h>
#include<string.h>

#define MAX 501
#define MAXP  1030

int point[MAXP],deg[MAX],G[MAX][MAX];
int n=0;


void euler(int u)
{
	int v;
	for(v=1;v<MAX;++v)
	{
		if(G[u][v])
		{
			G[u][v]--;
			G[v][u]--;
			euler(v);
			point[n++]=v;
		}
	}
}


int main(){
	FILE *fin=fopen("fence.in","r");
	FILE *fout=fopen("fence.out","w");

	int F,i,a,b;
	int start;


	fscanf(fin,"%d",&F);    //输入
	
	for(i=0;i<F;++i)
	{
		fscanf(fin,"%d%d",&a,&b);
		G[a][b]++;    //两个点之间可能有多条边
		G[b][a]++;
		deg[a]++;
		deg[b]++;
	}



	start=1;
	for(i=1;i<MAX;++i)     //寻找起点
		if(deg[i]%2)
		{
			start=i;
			break;
		}


	euler(start);   //寻找欧拉路径
	point[F]=start;


	for(i=F;i>=0;--i)   //输出
		fprintf(fout,"%d\n",point[i]);
	

	return 0;
}