安利：IOI2020題解-洛谷部落格 IOI2020題解-cnblogs

題目連結-LOJ 題目連結-洛谷

首先，如果存在 \(p_{i,j}=0\) ,那麼就分成了多個聯通塊，這些聯通塊分開做即可。

由於 $p_{i,j} \leq 3 $ 所以一個聯通塊當中不能出現多於一個環，否則必然會有兩個點之間有 \(4\) 條路徑。

沒有環的情況(聯通塊內 \(p_{i,j}\) 均為 \(1\) )先判掉，那麼這個聯通塊正好有一個環。

不難發現，如果兩個點 \(x\) 和 \(y\) 之間的路徑需要經過多於一個環上的節點，那麼 \(p_{x,y}\) 就等於 \(2\) ,否則 \(p_{x,y}\) 為 \(1\)

把所有的 \(p_{x,y} = 1\) 的 \((x,y)\) 合併到一個聯通塊中，然後把這些塊之間串成一個環，然後 \(check\) 是否合法即可。

\(\Theta (n^2)\)

code(LOJ上通過):

#include <bits/stdc++.h>
#include "supertrees.h"
using namespace std;
const int N = 1001;
int n,cur[N],cntc,G[N][N],ans[N][N];
int p[N],l,fa[N];
inline int Find(int x){ return x == fa[x] ? x : (fa[x] = Find(fa[x])); }
inline void Merge(int x,int y){ fa[Find(x)] = Find(y); }
inline void add_edge(int x,int y){ ans[x][y] = ans[y][x] = 1; }
inline bool solve(int c){
	int i,j; bool is = 0;
	for (l = 0,i = 1; i <= n; ++i) if (cur[i] == c) p[++l] = i;
	for (i = 1; i <= l; ++i) for (j = 1; j <= l; ++j) if (G[p[i]][p[j]] != 1) is = 1;
	if (l == 1) return 1;
	if (l == 2){ if (G[p[1]][p[2]] > 1) return 0; add_edge(p[1],p[2]); return 1; }
	if (!is){ for (i = 1; i < l; ++i) add_edge(p[i],p[i+1]); return 1; }
	for (i = 1; i <= l; ++i) fa[i] = i;
	for (i = 1; i <= l; ++i) for (j = 1; j <= l; ++j) if (G[p[i]][p[j]] == 1) Merge(i,j);
	for (i = 1; i <= l; ++i) fa[i] = Find(i);
	for (i = 1; i <= l; ++i) for (j = 1; j <= l; ++j)
		if (fa[i] == fa[j] && G[p[i]][p[j]] == 2) return 0;
	for (i = 1; i <= l; ++i) for (j = 1; j <= l; ++j)
		if (fa[i] != fa[j] && G[p[i]][p[j]] == 1) return 0;
	int x = 0,lst = 0,fir,cnt = 0;
	for (i = 1; i <= l; ++i){
		x = is = 0;
		for (j = 1; j <= l; ++j) if (fa[j] == i) is = 1;
		if (!is) continue;
		++cnt;
		for (j = 1; j <= l; ++j) if (j != i && fa[j] == i){
			if (x) add_edge(p[x],p[j]); x = j;
		}
		if (x) add_edge(p[x],p[i]);
		if (lst) add_edge(p[lst],p[i]); else fir = i; lst = i;
	}
	if (cnt < 3) return 0;
	add_edge(p[lst],p[fir]);
	return 1;
}
inline void getcur(int x,int c){
	cur[x] = c;
	for (int i = 1; i <= n; ++i) if (x != i && G[x][i] && !cur[i]) getcur(i,c);
}
inline int MAIN(){
	int i,j;
	for (i = 1; i <= n; ++i) if (!cur[i]) getcur(i,++cntc);
	for (i = 1; i <= n; ++i) for (j = 1; j <= n; ++j) if (cur[i] != cur[j] && G[i][j]) return 0;
	for (i = 1; i <= n; ++i) for (j = 1; j <= n; ++j) if (cur[i] == cur[j] && !G[i][j]) return 0;
	for (i = 1; i <= cntc; ++i) if (!solve(i)) return 0;
	vector<vector<int> >Ans; vector<int>ret; Ans.clear();
	for (i = 1; i <= n; ++i){
		ret.resize(n);
		for (j = 0; j < n; ++j) ret[j] = ans[i][j+1],cerr << ans[i][j+1] << ' '; cerr<<'\n';
		Ans.push_back(ret);
	}
	build(Ans);
	return 1;
}
int construct(vector<vector<int> > p){
	int i,j; n = p[0].size();
	for (i = 0; i < n; ++i) for (j = 0; j < n; ++j) if (p[i][j] == 3) return 0;
	for (i = 1; i <= n; ++i) for (j = 1; j <= n; ++j) G[i][j] = p[i-1][j-1];
	return MAIN();
}
/*
int main(){
	int n,i,j;
	cin >> n;
	vector<vector<int> >p; vector<int>q; p.clear(),q.resize(n);
	for (i = 1; i <= n; ++i){
		for (j = 0; j < n; ++j) cin >> q[j];
		p.push_back(q);
		
	}
	cout << construct(p) << '\n';
}
*/