Codeforces Beta Round #29 (Div. 2, Codeforces format) C. Mail Stamps 离散化拓扑排序

C. Mail Stamps

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/problemset/problem/29/C

Description

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.

Sample Input

2
1 100
100 2

Sample Output

2 100 1

HINT

题意

给你一条链，让你从头输出到尾

题解：

离散化一下，然后在跑一发拓扑排序就好了

代码

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 2000001
#define mod 1000000007
#define eps 1e-9
int Num;
char CH[20];
const int inf=0x3f3f3f3f;
inline ll read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

//**************************************************************************************

int n;
vector<int> q;
map<int,int> H;
map<int,int> h;
int a[maxn];
int b[maxn];
vector<int> e[maxn];
int d[maxn];
int vis[maxn];
int main()
{
    n=read();
    for(int i=0;i<n;i++)
    {
        a[i]=read();
        b[i]=read();
        q.push_back(a[i]);
        q.push_back(b[i]);
    }
    sort(q.begin(),q.end());
    q.erase(unique(q.begin(),q.end()),q.end());
    for(int i=0;i<q.size();i++)
        H[q[i]]=i,h[i]=q[i];
    for(int i=0;i<n;i++)
    {
        e[H[a[i]]].push_back(H[b[i]]);
        e[H[b[i]]].push_back(H[a[i]]);
        d[H[a[i]]]++;
        d[H[b[i]]]++;
    }
    int flag=1;
    queue<int> qq;
    for(int i=0;i<q.size();i++)
    {
        if(d[H[q[i]]]==1)
        {
            qq.push(H[q[i]]);
            vis[H[q[i]]]=1;
            break;
        }
    }
    while(!qq.empty())
    {
        int v=qq.front();
        printf("%d ",h[v]);
        vis[v]=1;
        qq.pop();
        for(int i=0;i<e[v].size();i++)
        {
            if(vis[e[v][i]])
                continue;
            d[e[v][i]]--;
            if(d[e[v][i]]<=1)
                qq.push(e[v][i]);
        }
    }
}