BZOJ4727 [POI2017]Turysta 【競賽圖哈密頓路徑/迴路】

#題目連接 BZOJ4727 #題解 前置芝士

1.競賽圖存在哈密頓路徑 2.競賽圖存在哈密頓迴路，當且僅當它是強聯通的

因此咱們將圖縮點後，拓撲排序後必定是一條鏈，且以前的塊內的點和以後塊內的點的邊必定全都由前面指向後面 而每一個塊都是強聯通的，因此咱們從起點出發，必定能找到一條路徑走完後面全部點

咱們只需預處理出每一個強聯通塊內的一條哈密頓迴路，就能夠求出答案了

如今問題轉化成了求競賽圖的哈密頓迴路 咱們先求出一條哈密頓路徑

##哈密頓路徑 從競賽圖中任意一個點出發向外擴展，維護一個鏈表 倘若擴展到點$u$ 1.若是$u$指向鏈頭或鏈尾，直接加入鏈表 2.不然鏈的中間必定存在相鄰兩點，使得$i$指向$u$，$u$指向$i + 1$，這時候把$u$插入之間便可

##哈密頓迴路 咱們在哈密頓路徑的基礎上構造哈密頓迴路 首先若是存在如圖狀況，前$4$個點構成迴路 咱們先找到最大的一個這樣的迴路，而後只需處理後面不在圈內的幾個點

對於一個點$u$，若是存在一條$u$指向圈內點的邊，那麼$u$能夠插入圈內 不然跳過$u$，將$u$和以後插入圈內的點一塊兒插入圈內 因爲圖是強聯通的，因此最後必定能所有加入

複雜度$O(n^2)$

#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;
}