hdu多校2C

题意:找多条路径覆盖所有的边,求最小路径数,要求输出路径
题解:新建一个点n+1,所有奇点向它连边,然后跑欧拉回路,最后把新加的边删去,一段连续的边就是一条路径
= =但是由于太久没写欧拉回路以及之前的板子有点问题,导致比赛时没做出来
还有一个坑点是,明明题目里说了没有重边,我规定不能访问父亲居然就wa了!!!!

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=200000+10,maxn=400000+10,inf=0x3f3f3f3f;

struct edge{
    int to,Next,id;
}e[maxn];

int cur[N];
int cnt,head[N],deg[N];
vi s[N];
int res,n,m,sum;
bool ban[maxn];

inline void init()
{
    cnt=res=sum=0;
    for(int i = 1; i <= n + 1; i++) {
        head[i] = -1;
        deg[i] = 0;
    }
}

inline void add(int u,int v,int id)
{
    e[cnt].to=v;
    e[cnt].Next=head[u];
    e[cnt].id=id;
    ban[cnt] = false;
    head[u]=cnt++;
    e[cnt].to=u;
    e[cnt].Next=head[v];
    e[cnt].id=id;
    ban[cnt] = false;
    head[v]=cnt++;
}

void dfs(int u,int f)
{
//    cout << u << " " << f << endl;
//    for(int i=head[u];~i;i=e[i].Next)
    while(head[u] != -1)
    {
        int i = head[u];
        int x=e[i].to;
        head[u] = e[i].Next;
        if(!ban[e[i].id])
        {
            ban[e[i].id]=1;
            dfs(x,u);
            if(u!=n+1)
            {
                if(x!=n+1)
                {
                    s[res].pb(i&1?-e[i].id:e[i].id);
                }
                else
                {
                    if((int)s[res].size()!=0)res++;
                }
            }
        }
    }
}

int main()
{
    int a, b;
    while(~scanf("%d%d",&n, &m))
    {
        init();
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d", &a, &b);
            add(a,b,i);
            deg[a]++,deg[b]++;
        }
        int te=m;
        for(int i=1;i<=n;i++)
            if(deg[i]&1)
                add(i,n+1,++te);
        dfs(n+1,-1);
        if((int)s[res].size()!=0)res++;
            for(int i=1,j;i<=n;i++)
                while((j=head[i])!=-1){
                    head[i] = e[j].Next;
                    if(!ban[e[j].id])
                    {
                        dfs(i,-1);
                        if(s[res].size()!=0)res++;
                    }
                }
        printf("%d\n",res);
        for(int i=0;i<res;i++)
        {
            printf("%d",s[i].size());
            for(int j=s[i].size() - 1;j >= 0;--j)
                printf(" %d",s[i][j]);
            puts("");
            s[i].clear();
        }

//        for(int i = 1; i <= n; i++) printf("%d ", head[i]);
    }
    return 0;
}
/********************
6 6
1 2
2 3
3 4
2 5
3 5
2 6

6 6
1 2
2 3
1 3
4 5
5 6
4 6

31 36
12 28
12 27
27 28
15 26
15 25
25 26
17 23
17 24
23 24
20 21
20 22
21 22
29 30
30 31
29 31
1 2
2 3
3 4
1 4
4 5
4 6
5 7
6 7
7 8
8 9
9 10
10 11
11 12
10 14
14 15
10 13
13 16
16 17
13 18
18 19
19 20

8 7
1 2
2 3
3 4
4 8
5 6
6 7
7 4

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