昨天UNR考完之后，有许多小朋友问我标程怎么写的owo

那我挂一下好了

//Awwawa! Dis cold yis ratten buy tEMMIE!
#include <bits/stdc++.h>
#define ll long long
#define maxn 100005 /*rem*/
#define mod 998244353
#define db double
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define pi pair<int, int>
#define fi first
#define se second

template <typename T> bool chkmax(T &x,T y){return x<y?x=y,true:false;}
template <typename T> bool chkmin(T &x,T y){return x>y?x=y,true:false;}

using namespace std;
ll ksm(ll a, ll b) {
   if (!b) return 1;
   ll ns = ksm(a, b >> 1);
   ns = ns * ns % mod;
   if (b & 1) ns = ns * a % mod;
   return ns;
}
int out[maxn], in[maxn];
vector<pi> eg;
mt19937 x;
int gt(int a, int b) {
    if (a == b) return 1;
    if (!a) return 0;
    return gt(out[a], b);
}
int main() {
    for (int t = 1/*1*/; t <= 10; t++) {
        char nm[30];
        sprintf(nm, "hamil%d.in", t);
        ifstream rd(nm);
        sprintf(nm, "hamil%d.ans", t);
        cout << "WORK" << t << endl;
        ofstream ot(nm);
        int n, m;
        rd >> n >> m;
        eg.clear();
        for (int i = 1; i <= m; i++) {
            int u, v;
            rd >> u >> v;
            eg.pb(mp(u, v));
        }
        shuffle(eg.begin(), eg.end(), x);
        while (1) {
            int tot = 0, cnt = 0;
            memset(out, 0, sizeof(out));
            memset(in, 0, sizeof(in));
            while (1) {
                int fl = 0;
                shuffle(eg.begin(), eg.end(), x);
                for (auto v : eg) {
                    if (in[v.se] && out[v.fi]) continue;
                    if (gt(v.se, v.fi)) continue;
                    if (in[v.se] || out[v.fi]) 
                        if (x() & 1) continue;
                    if (!in[v.se] && !out[v.fi])
                        tot++, fl = 1;
                    if (in[v.se]) {
                        out[in[v.se]] = 0;
                        in[v.se] = 0;
                    }
                    if (out[v.fi]) {
                        in[out[v.fi]] = 0;
                        out[v.fi] = 0;
                    }
                    in[v.se] = v.fi;
                    out[v.fi] = v.se;
                }
                if (tot == n - 1) break;
                if (fl) cnt = 0;
                else {
                    cnt++;
                    if (cnt >= 3000000) break;
                }
                if (x() % 20 == 0)cout << tot << endl;
            }
            if (tot == n - 1) break;
            cout << "AGAIN" << endl;
        }
        for (int i = 1; i <= n; i++) 
            if (!in[i]) {
                int pl = i;
                while (pl)
                    ot << pl << ' ', 
                    pl = out[pl];
                break;
            } 
        ot << endl;
    }
    return 0;
}

这是当时的标程w 由于我很懒，什么优化也没加

今天空闲时拷了个LCT的板子，又封装了一下，现在可以在4s跑完UNR4 D2T3

代码如下

//Awwawa! Dis cold yis ratten buy tEMMIE!
#include <bits/stdc++.h>
/*这是一份UNR4 D2T3示例代码 只需要几秒就能AC该题
hamil::work是封装好的函数，不需初始化可直接调用
其参数为: int n, vector<pair<int, int> > edges, int mx_ch = -1
其中n为有向图点数（从1开始标号），edges为有向边的集合（edges中元素pair<u, v>表示从u到v有向边），mx_ch为最大调整次数。如果不设初值，默认为(n+100)*(n+50)
这个函数时间复杂度不超过mx_ch * log(n) 如果能找到哈密顿链，往往远快于这个上界
如果有哈密顿链，函数有较大概率返回一条。链经过的节点按照经过的顺序储存在返回的vector<int> 中。
如果函数未能找到，会返回空的vector<int>。这时可以调大mx_ch阈值，或者多试几次w
欢迎大家喂各种图调戏它x
*/

using namespace std;
namespace hamil {
    template <typename T> bool chkmax(T &x,T y){return x<y?x=y,true:false;}
    template <typename T> bool chkmin(T &x,T y){return x>y?x=y,true:false;}
    #define vi vector<int>
    #define pb push_back
    #define mp make_pair
    #define pi pair<int, int>
    #define fi first
    #define se second
    #define ll long long
    namespace LCT {
        vector<vi> ch;
        vi fa, rev;
        void init(int n) {
            ch.resize(n + 1);
            fa.resize(n + 1);
            rev.resize(n + 1);
            for (int i = 0; i <= n; i++)
                ch[i].resize(2), 
                ch[i][0] = ch[i][1] = fa[i] = rev[i] = 0;
        }
        bool isr(int a)
        {
            return !(ch[fa[a]][0] == a || ch[fa[a]][1] == a);
        } 
        void pushdown(int a)
        {
            if(rev[a])
            {
                rev[ch[a][0]] ^= 1, rev[ch[a][1]] ^= 1;
                swap(ch[a][0], ch[a][1]);
                rev[a] = 0;
            }
        }
        void push(int a)
        {
            if(!isr(a)) push(fa[a]);
            pushdown(a); 
        }
        void rotate(int a)
        {
            int f = fa[a], gf = fa[f];
            int tp = ch[f][1] == a;
            int son = ch[a][tp ^ 1];
            if(!isr(f)) 
                ch[gf][ch[gf][1] == f] = a;    
            fa[a] = gf;

            ch[f][tp] = son;
            if(son) fa[son] = f;

            ch[a][tp ^ 1] = f, fa[f] = a;
        }
        void splay(int a)
        {
            push(a);
            while(!isr(a))
            {
                int f = fa[a], gf = fa[f];
                if(isr(f)) rotate(a);
                else
                {
                    int t1 = ch[gf][1] == f, t2 = ch[f][1] == a;
                    if(t1 == t2) rotate(f), rotate(a);
                    else rotate(a), rotate(a);    
                } 
            } 
        }
        void access(int a)
        {
            int pr = a;
            splay(a);
            ch[a][1] = 0;
            while(1)
            {
                if(!fa[a]) break; 
                int u = fa[a];
                splay(u);
                ch[u][1] = a;
                a = u;
            }
            splay(pr);
        }
        void makeroot(int a)
        {
            access(a);
            rev[a] ^= 1;
        }
        void link(int a, int b)
        {
            makeroot(a);
            fa[a] = b;
        }
        void cut(int a, int b)
        {
            makeroot(a);
            access(b);
            fa[a] = 0, ch[b][0] = 0;
        }
        int fdr(int a)
        {
            access(a);
            while(1)
            {
                pushdown(a);
                if (ch[a][0]) a = ch[a][0];
                else {
                    splay(a);
                    return a;
                }
            }
        }
    }
    vi out, in;
    vi work(int n, vector<pi> eg, ll mx_ch = -1) { // mx_ch : 最大调整次数  如果不设初值，默认为(n + 100) * (n + 50) 
        // 如果存在，有较大概率返回一条路径 
        // 如果失败 返回0 
        out.resize(n + 1), in.resize(n + 1);
        LCT::init(n);
        for (int i = 0; i <= n; i++) in[i] = out[i] = 0;
        if (mx_ch == -1) mx_ch = 1ll * (n + 100) * (n + 50); //时间上限 
        vector<vi> from(n + 1), to(n + 1);
        for (auto v : eg)
            from[v.fi].pb(v.se), 
            to[v.se].pb(v.fi);
        unordered_set<int> canin, canout;
        for (int i = 1; i <= n; i++)
            canin.insert(i), 
            canout.insert(i); 
        mt19937 x(time(0));
        int tot = 0;
        while (mx_ch >= 0) {
        //    cout << tot << ' ' << mx_ch << endl;
            vector<pi> eg;
            for (auto v : canout)
                for (auto s : from[v])
                    if (in[s] == 0) {
                        assert(canin.count(s));
                        continue;
                    }
                    else eg.pb(mp(v, s));
            for (auto v : canin)
                for (auto s : to[v])
                    eg.pb(mp(s, v));
            shuffle(eg.begin(), eg.end(), x);
            if (eg.size() == 0) break;
            for (auto v : eg) {
                mx_ch--;
                if (in[v.se] && out[v.fi]) continue;
                if (LCT::fdr(v.fi) == LCT::fdr(v.se)) continue;
                if (in[v.se] || out[v.fi]) 
                    if (x() & 1) continue;
                if (!in[v.se] && !out[v.fi]) 
                    tot++;
                if (in[v.se]) {
                    LCT::cut(in[v.se], v.se);
                    canin.insert(v.se);
                    canout.insert(in[v.se]);
                    out[in[v.se]] = 0;
                    in[v.se] = 0;
                }
                if (out[v.fi]) {
                    LCT::cut(v.fi, out[v.fi]);
                    canin.insert(out[v.fi]);
                    canout.insert(v.fi);
                    in[out[v.fi]] = 0;
                    out[v.fi] = 0;
                }
                LCT::link(v.fi, v.se);
                canin.erase(v.se);
                canout.erase(v.fi);
                in[v.se] = v.fi;
                out[v.fi] = v.se;
            }
            if (tot == n - 1) {
                vi cur;
                for (int i = 1; i <= n; i++) 
                    if (!in[i]) {
                        int pl = i;
                        while (pl) {
                            cur.pb(pl), 
                            pl = out[pl];
                        }
                        break;
                    } 
                return cur;
            }
        }
        //fail to find a path
        return vi();
    }
}
int main() {
    for (int t = 1; t <= 10; t++) {
        char nm[30];
        sprintf(nm, "hamil%d.in", t);
        ifstream rd(nm);
        sprintf(nm, "hamil%d.out", t);
        cout << "WORK" << t << endl;
        ofstream ot(nm);
        int n, m;
        rd >> n >> m;
        vector<pi> eg;
        for (int i = 1; i <= m; i++) {
            int u, v;
            rd >> u >> v;
            eg.pb(mp(u, v));
        }
        l1: 
        vi ed = hamil::work(n, eg);
        if (!ed.size()) goto l1;
        ot << ed.size() << endl;
        for (auto v : ed)
            ot << v << ' ';
        ot << endl;
    }
    return 0;
}

一些附加说明：

如果给定起点终点，只需新建两个点$U,V$ ； 然后 $U$ 到起点连一条边，终点到 $V$ 连一条边

如果要求哈密顿圈，可以枚举一条边，转化成给定起点终点情形。

如果是无向图，只需正反边各加一遍即可。