Codeforces 29C：Mail Stamps（STL的应用+DFS）

C. Mail Stamps

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

One day Bob got a letter in an envelope. Bob knows that when Berland's post officers send a letter directly from city «A» to city «B», they stamp it with «A B», or «B A». Unfortunately, often it is impossible to send a letter directly from the city of the sender to the city of the receiver, that's why the letter is sent via some intermediate cities. Post officers never send a letter in such a way that the route of this letter contains some city more than once. Bob is sure that the post officers stamp the letters accurately.

There are n stamps on the envelope of Bob's letter. He understands that the possible routes of this letter are only two. But the stamps are numerous, and Bob can't determine himself none of these routes. That's why he asks you to help him. Find one of the possible routes of the letter.

Input

The first line contains integer n (1 ≤ n ≤ 105) — amount of mail stamps on the envelope. Then there follow n lines with two integers each — description of the stamps. Each stamp is described with indexes of the cities between which a letter is sent. The indexes of cities are integers from 1 to 109. Indexes of all the cities are different. Every time the letter is sent from one city to another, exactly one stamp is put on the envelope. It is guaranteed that the given stamps correspond to some valid route from some city to some other city.

Output

Output n + 1 numbers — indexes of cities in one of the two possible routes of the letter.

Examples

input

21 100100 2

output

2 100 1 

input

33 1100 23 2

output

100 2 3 1

题目大意：给你写连接的点的信息，问走完这些点的路径为多少，输出路径，保证一定有路径，且每个点只经过一次。

解题思路：这里点的标号太大，考虑离散化处理这些点，我这里用的map，应该也能用结构题写吧。映射过之后，建立链表，然后从边的点开始扫，扫到底部开始倒着输出。

代码如下：

<pre name="code" class="cpp">#include <cstdio>#include <algorithm>#include <vector>#include <cstring>#include <map>using namespace std;map<int,int>duiying;//离散化用的比如map[99999999][1]，就用1来代表那个99999999 vector<int>G[100010];//储存链表 int f[100010];//f[1]=99999999，编号为1的数离散化之前本来是99999999 int vis[100010];//标记是否访问过该节点 int num;//看总共有多少个不同的数 bool dfs(int dian,int step){if(step==num-1)//扫到头了，那么倒着输出 {printf("%d",f[dian]);}for(int i=0;i<G[dian].size();i++)//vis的目的是防止他往回搜索 {if(vis[G[dian][i]]==0)//看他连接的节点是否能访问 {vis[G[dian][i]]=1;//能访问则标记已经访问 dfs(G[dian][i],step+1);//接着找他连接的点所能连接的点 printf(" %d",f[dian]);}}}int main(){int n;scanf("%d",&n);for(int i=0;i<100010;i++)//记得初始化清空 {G[i].clear();}duiying.clear();//map这样清零 num=1;//初始化 for(int i=0;i<n;i++){int a,b;scanf("%d%d",&a,&b);if(duiying[a]==0)//没出现过a这个数，那么给他映射下 {duiying[a]=num;//把大的a离散化为较小的num f[num]=a;//储存原来的值 num++;}if(duiying[b]==0)//同上 {duiying[b]=num;f[num]=b;num++;}G[duiying[a]].push_back(duiying[b]);//建立联系 ，表示a连有b G[duiying[b]].push_back(duiying[a]);//建立联系 ，表示b连有a }int start;//从边上的点出发 for(int i=1;i<num;i++){if(G[i].size()==1)//找到最边上的点 {start=i;break;}}memset(vis,0,sizeof(vis));vis[start]=1;//标记一下 dfs(start,1);//第一个参数是出发点，第二个参数是扫过的节点数 printf("\n");return 0;}