【codeforces】29C—Mail Stamps

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

2
1 100
100 2

output

2 100 1

input

3
3 1
100 2
3 2

output

100 2 3 1

真心好题，用到了vector容器、map映射等STL知识，还有dfs算法，离散化思想以及各种小技巧和需要注意的细节。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
const int N = 1e5+10;
vector<int>graph
;//使用vector容器，相当于邻接表的作用，比邻接矩阵要快的多，但比邻接表慢点。
map<int,int>m;//使用映射实现离散化，因为城市的标号太大，存储起来占内存
int f
; //存储原值
bool vis
;
int n;
int dfs(int cur,int step) {
if(step==n+1) {
printf("%d",f[cur]);
return 1;
}
for(int i=0; i<graph[cur].size(); i++) {
if(vis[graph[cur][i]]==0) {
vis[graph[cur][i]]=1;
if(dfs(graph[cur][i],step+1)) {
printf(" %d",f[cur]);
return 1;
}
}
vis[graph[cur][i]]=0;// 回溯
}
}
int main() {
while(scanf("%d",&n)!=EOF) {
int a,b,pos=0;
for(int i=0; i<=N; i++) {
graph[i].clear();
}
memset(vis,0,sizeof(vis));
m.clear();
for(int i=0; i<n; i++) {
scanf("%d%d",&a,&b);
if(!m[a]) m[a]=++pos,f[pos]=a;
if(!m[b]) m[b]=++pos,f[pos]=b;
graph[m[a]].push_back(m[b]);
graph[m[b]].push_back(m[a]);
}
int start=1;
for(int i=1; i<=n+1; i++) {
if(graph[i].size()==1) {
start=i;
break;
}
}
vis[start]=1;
dfs(start,1);
printf("\n");
}
return 0;
}题目地址：http://codeforces.com/problemset/problem/29/C