hdu 2230 watchcow 【图论-欧拉回路-遍历-输出路径】

                                        Watchcow        Time Limit: 3000MS      Memory Limit: 65536K    Special Judge

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…

题目大意：输出所给欧拉图的遍历路径

AC代码：

# include <cstdio># include <cstring># define MAXN 100005# define MAXM 2000005struct EDGE{    int to;    int next;}edge[MAXM];int tot;int head[MAXN];int vis[MAXN];void Init(){    tot = 0;    memset(head, -1, sizeof(head));    memset(vis, 0, sizeof(vis));}void Addedge(int u, int v) //邻接表存放图{    edge[tot].to = v;    edge[tot].next = head[u];    head[u] = tot++;    edge[tot].to = u;    edge[tot].next = head[v];    head[v] = tot++;}void Dfs(int u){    for (int i = head[u]; i + 1; i = edge[i].next)    {        int v = edge[i].to;        if (!vis[i])        {            vis[i] = 1;            Dfs(v);        }    }    printf("%d\n", u);}int main(void){    int n, m;    while (~scanf("%d %d", &n, &m))    {        Init();        while (m--)        {            int u, v;            scanf("%d %d", &u, &v);            Addedge(u, v);        }        Dfs(1);    }    return 0;}