POJ22230 Watchcow (欧拉回路)

Watchcow

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 6477		Accepted: 2823		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...

收获：了解了欧拉回路

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
using namespace std;

const int INF=0x3f3f3f3f;
const double eps=1e-;
const double PI=acos(-1.0);
#define maxn 100050
int n, m, num;
int head[maxn], vis[maxn];
struct Edge
{
    int u, v, next;
};
Edge edge[maxn];
void addedge(int u, int v)
{
    edge[num].u = u;
    edge[num].v = v;
    edge[num].next = head[u];
    head[u] = num++;
}
int ans[maxn];
int cnt;
void dfs(int now)
{
    for(int i = head[now]; i != -; i = edge[i].next)
    {
        if(!vis[i])
        {
            vis[i] = ;
            //vis[i^1] = 1;
            dfs(edge[i].v);
            //ans[cnt++] = edge[i].v;
        }
    }
    printf("%d\n", now);
}
int main()
{
    while(~scanf("%d%d", &n, &m))
    {
        int u, v;
        num = ;
        memset(head, -, sizeof head);
        for(int i = ; i < m; i++)
        {
            scanf("%d%d", &u, &v);
            addedge(u, v);
            addedge(v, u);
        }
        cnt = ;
        memset(vis, , sizeof vis);
        dfs();
        for(int i = ; i < cnt; i++)
            printf("%d\n", ans[i]);
    }
    return ;
}