题目链接
题解
前置芝士
所以我们将图缩点后，拓扑排序后一定是一条链，且之前的块内的点和之后块内的点的边一定全都由前面指向后面
而每个块都是强联通的，所以我们从起点出发，一定能找到一条路径走完后面所有点
我们只需预处理出每个强联通块内的一条哈密顿回路，就可以求出答案了
现在问题转化成了求竞赛图的哈密顿回路
我们先求出一条哈密顿路径
哈密顿路径
从竞赛图中任意一个点出发向外扩展，维护一个链表
假若扩展到点\(u\)
1.如果\(u\)指向链头或链尾，直接加入链表
2.否则链的中间一定存在相邻两点，使得\(i\)指向\(u\)，\(u\)指向\(i + 1\)，这时候把\(u\)插入之间即可
哈密顿回路
我们在哈密顿路径的基础上构造哈密顿回路
首先如果存在如图情况，前\(4\)个点构成回路
我们先找到最大的一个这样的回路，然后只需处理后面不在圈内的几个点
对于一个点\(u\)，如果存在一条\(u\)指向圈内点的边，那么\(u\)可以插入圈内
否则跳过\(u\)，将\(u\)和之后插入圈内的点一起插入圈内
由于图是强联通的，所以最后一定能全部加入
复杂度\(O(n^2)\)
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<map>
#define LL long long int
#define REP(i,n) for (register int i = 1; i <= (n); i++)
#define cls(s,v) memset(s,v,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
#define res register
using namespace std;
const int maxn = 2005,maxm = 100005,INF = 0x3f3f3f3f;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}
return flag ? out : -out;
}
int tmp[20],ti;
inline void write(int x){
ti = 0;
while (x) tmp[++ti] = x % 10,x /= 10;
while (ti) putchar('0' + tmp[ti--]);
}
vector<int> S[maxn];
int n,G[maxn][maxn];
int dfn[maxn],low[maxn],Scc[maxn],st[maxn],top,cnt,scci;
int nxt[maxn],head[maxn],tail[maxn];
int Nxt[maxn],Head[maxn],Tail[maxn];
int g[maxn][maxn],de[maxn],q[maxn],hh,tt;
int tp[maxn],pos[maxn],ans[maxn],ansi,tot;
int c[maxn],ci;
void dfs(int u){
dfn[u] = low[u] = ++cnt;
st[++top] = u;
for (res int to = 1; to <= n; to++){
if (!G[u][to]) continue;
if (!dfn[to]){
dfs(to);
low[u] = min(low[u],low[to]);
}
else if (!Scc[to]) low[u] = min(low[u],dfn[to]);
}
if (low[u] == dfn[u]){
scci++;
do{
Scc[st[top]] = scci;
S[scci].push_back(st[top]);
}while (st[top--] != u);
}
}
void workline(){
int siz;
for (res int p = 1; p <= scci; p++){
siz = S[p].size();
head[p] = tail[p] = S[p][0];
for (res int i = 1; i < siz; i++){
int u = S[p][i];
if (G[u][head[p]]) nxt[u] = head[p],head[p] = u;
else if (G[tail[p]][u]) nxt[tail[p]] = u,tail[p] = u;
else for (res int j = head[p]; j; j = nxt[j])
if (G[j][u] && G[u][nxt[j]]){
nxt[u] = nxt[j]; nxt[j] = u; break;
}
}
}
}
void workcir(){
int p,last;
for (int i = 1; i <= scci; i++){
ci = 0; p = 1;
for (int k = head[i]; k; k = nxt[k]) c[++ci] = k;
for (int k = ci; k; k--)
if (G[c[k]][head[i]]) {p = k; break;}
Head[i] = c[1]; Tail[i] = c[p];
for (int k = 1; k < p; k++) Nxt[c[k]] = c[k + 1];
last = c[p + 1];
for (int k = p + 1; k <= ci; k++){
int u = c[k],flag = false;
for (int j = Nxt[Head[i]],pre = Head[i]; j; j = Nxt[pre = j])
if (G[u][j]){
Nxt[pre] = last;
Nxt[u] = j;
flag = true;
break;
}
if (flag) last = c[k + 1];
else Nxt[u] = c[k + 1];
}
Nxt[Tail[i]] = Head[i];
}
}
void work(){
for (res int i = 1; i <= n; i++){
int u = Scc[i];
for (res int j = 1; j <= n; j++)
if (G[i][j] && Scc[j] != u && !g[u][Scc[j]])
de[Scc[j]]++,g[u][Scc[j]] = 1;
}
for (res int i = 1; i <= scci; i++) if (!de[i]) q[++tt] = i;
int u; hh = 1;
while (hh <= tt){
u = q[hh++]; pos[u] = ++tot; tp[tot] = u;
for (int i = 1; i <= scci; i++) if (g[u][i]){
if (!(--de[i])) q[++tt] = i;
}
}
for (res int u = 1; u <= n; u++){
int s = Scc[u];
ans[ansi = 1] = u;
for (res int i = Nxt[u]; i != u; i = Nxt[i]) ans[++ansi] = i;
for (res int j = pos[s] + 1; j <= scci; j++){
int t = tp[j];
ans[++ansi] = Head[t];
for (res int i = Nxt[Head[t]]; i != Head[t]; i = Nxt[i])
ans[++ansi] = i;
}
write(ansi); putchar(' ');
for (res int i = 1; i <= ansi; i++){
write(ans[i]);
if (i < ansi) putchar(' ');
}
puts("");
}
}
int main(){
n = read();
for (res int i = 2; i <= n; i++)
for (res int j = 1; j < i; j++)
G[i][j] = ((G[j][i] = read()) ^ 1);
REP(i,n) if (!dfn[i]) dfs(i);
workline();
//puts("LXT");
workcir();
//puts("LXT");
work();
return 0;
}