POI[2011]Garbage 欧拉回路的正确姿势

题目大意是讲给定一张图，每跳边有编号$0$或$1$，每次可以将一个简单环上所有边$\oplus 1$，构造方案使得每跳边的编号是某个值，方案中涉及的边不超过5m

如果方案中的两个环有相交部分，相交部分没有变化，去掉之后2个环合并成了一个环。所以存在一种方案使得所有环都不相交

于是删除需要进过偶数次的边，对每个联通块求欧拉回路（不是道路），然后再用栈搞一搞弄出所有的环，时间复杂度$O(n + m)$

TLE...，为了卡常数还怒写人工栈...然后发现欧拉回路部分，一开始我是这么写的

void dfs(int u)
{
    vis[u] = 1;
    for (int i = head[u], v; v = e[i].node, i; i = e[i].next)
        if (!flag[i])
        {
            flag[i] = flag[i ^ 1] = 1;
            dfs(v);
            mem[++mem[0]] = i;
        }
}

一直都以为这样写就是对的，然后发现自己太naive了，每个点都从头开始找没有被标记过的边，复杂度是能退化的。。

正确的姿势应该是记录每个点最后一条标记的边是哪一条，然后从那一条边开始继续扫描（人工栈版本）

void dfs(int rt)
{
    S[top = 1].u = rt, S[1].i = head[rt], S[1].t = 0;
    while (top)
    {
        int u = S[top].u, i = S[top].i, tmp = top; S[top].t++; 
        if (S[top].t == 2) {mem[++mem[0]] = i; S[top].t = 1;}
        i = cur[u];
        vis[u] = 1;
        for (int v; v = e[i].node, i; i = e[i].next)
        if (!flag[i])
        {
            flag[i] = flag[i ^ 1] = 1;
            S[++top].u = v; S[top].i = cur[v]; 
            S[top].t = 0; S[tmp].i = i; cur[u] = i;
            break;
        }
        if (tmp == top) top--;
    }
}

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <ctime>
using namespace std;
#define rep(i, l, r) for (int i = l; i <= r; i++)
#define drep(i, r, l) for (int i = r; i >= l; i--)
typedef long long ll;
const int N = 1e5 + 8, M = 1e6 + 8;
int n, m, tot, head[N], mem[M << 1], top, sz, cnt[N], q[M << 1], deg[N], vis[N], cur[N];
bool flag[M << 1];
int ans[M * 5], len, sta[M];
int xgw;
//double t1, t2;
int getint()
{
    char c; int num = 0, w = 1;
    for (c = getchar(); !isdigit(c) && c != '-'; c = getchar());
    if (c == '-') c = getchar(), w = -1;
    for (;isdigit(c); c = getchar()) num = num * 10 + c - '0';
    return num * w;
}
struct Edge{int next, node;}e[M << 1];
inline void add(int x, int y)
{
    e[++tot].next = head[x], head[x] = tot, e[tot].node = y;
    e[++tot].next = head[y], head[y] = tot, e[tot].node = x;
}
struct RGZ
{
    int u, i, t;
}S[M];
void dfs(int rt)
{
    S[top = 1].u = rt, S[1].i = head[rt], S[1].t = 0;
    while (top)
    {
        int u = S[top].u, i = S[top].i, tmp = top; S[top].t++; 
        if (S[top].t == 2) {mem[++mem[0]] = i; S[top].t = 1;}
        i = cur[u];
        vis[u] = 1;
        for (int v; v = e[i].node, i; i = e[i].next)
        if (!flag[i])
        {
            flag[i] = flag[i ^ 1] = 1;
            S[++top].u = v; S[top].i = cur[v]; 
            S[top].t = 0; S[tmp].i = i; cur[u] = i;
            break;
        }
        if (tmp == top) top--;
    }
}
inline bool calc(int rt)
{
    mem[0] = 0; top = 0;
    dfs(rt);
    if (!mem[0]) return 1;
    q[top = 1] = e[mem[mem[0]] ^ 1].node; cnt[q[1]]++;
    drep(i, mem[0], 1)
    {
        int u = e[mem[i]].node;
        cnt[u]++; 
        if (cnt[u] == 2)
        {
            ++sz; ans[++len] = u; sta[sz] = len;
            while (cnt[u] > 1)
                ans[++len] = q[top], cnt[q[top]]--, top--;
        }
        q[++top] = u;
    }
    if (top > 1) return 0;
}
void solve()
{
    rep(i, 1, n) 
    if (deg[i] & 1)
    {
        printf("NIE\n");
        return;
    }
    rep(i, 1, n) cur[i] = head[i];
    rep(i, 1, n) 
        if (deg[i] && !vis[i]) 
            if (!calc(i))
            {
                printf("NIE\n");
                return;
            }
    printf("%d\n", sz); sta[sz + 1] = len + 1;
    rep(i, 1, sz)
    {
        printf("%d ", sta[i + 1] - sta[i] - 1);
        rep(j, sta[i], sta[i + 1] - 1) printf("%d ", ans[j]);
        putchar('\n');
    }
}
int main()
{
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    scanf("%d%d", &n, &m); tot = 1;
    rep(i, 1, m)
    {
        int u = getint(), v = getint(), x = getint(), y = getint();
        if (y != x) add(u, v), deg[u]++, deg[v]++;
    }
    solve();
    //fclose(stdin); fclose(stdout);
    return 0;
}