poj2230 歐拉回路

http://poj.org/problem?id=2230

Description

Bessie's been appointed the new watch-cow for the farm. Every night, it's her job to walk across the farm and make sure that no evildoers are doing any evil. She begins at the barn, makes her patrol, and then returns to the barn when she's done. 

If she were a more observant cow, she might be able to just walk each of M (1 <= M <= 50,000) bidirectional trails numbered 1..M between N (2 <= N <= 10,000) fields numbered 1..N on the farm once and be confident that she's seen everything she needs to see. But since she isn't, she wants to make sure she walks down each trail exactly twice. It's also important that her two trips along each trail be in opposite directions, so that she doesn't miss the same thing twice. 

A pair of fields might be connected by more than one trail. Find a path that Bessie can follow which will meet her requirements. Such a path is guaranteed to exist.

Input

* Line 1: Two integers, N and M. 

* Lines 2..M+1: Two integers denoting a pair of fields connected by a path.

Output

* Lines 1..2M+1: A list of fields she passes through, one per line, beginning and ending with the barn at field 1. If more than one solution is possible, output any solution.

Sample Input

4 5
1 2
1 4
2 3
2 4
3 4

Sample Output

1
2
3
4
2
1
4
3
2
4
1

Hint

OUTPUT DETAILS: 

Bessie starts at 1 (barn), goes to 2, then 3, etc...

 

題解：

一個圖是歐拉圖，那麼他的子圖也是一個歐拉圖，只需dfs便可

#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN=1e5+10;
struct node{
    int v;
    int next;
}G[MAXN];
int head[MAXN],cnt;
bool vis[MAXN];
int ans[MAXN];
void add(int u,int v)
{
    G[++cnt].v=v;
    G[cnt].next=head[u];
    head[u]=cnt;
}
int n,m,k=0;
int dfs(int u)
{
    for (int i = head[u]; i!=-1 ; i=G[i].next) {
        if(!vis[i])
        {
            vis[i]= true;
            dfs(G[i].v);
            ans[k++]=G[i].v;
        }
    }
}
int main() {
    while (scanf("%d%d",&n,&m)!=EOF)
    {
        cnt=0,k=0;
        memset(head,-1, sizeof(head));
        memset(vis,false, sizeof(vis));
        int u,v;
        for (int i = 0; i <m ; ++i) {
            scanf("%d%d",&u,&v);
            add(u,v);
            add(v,u);
        }
        dfs(1);
        for (int i = 0; i <k ; ++i) {
            printf("%d\n",ans[i]);
        }
        printf("1\n");
    }
    return 0;
}
//poj2230