(Un)labeled graphs 题解

orz jiangly

通信题。不禁让人想到。

对于这道题目，考虑要还原原来的点的编号，而题目条件里有一个 。

还是非常明显，发现我们可以搞出一排新的点让它们构成一条新的链。然后第 \(i\) 个点往原图中编号为 \(u\) 的点连边，满足 \(u \& 2^{i-1}\)。

现在我们需要解决的问题是如何辨认出这条链。发现还多了三个点可以供我们操作。设其为 A,B,C。

考虑怎么搞出一个特殊点让我们一开始就可以辨认出来。令其度数最大即可。于是我们让 A 与其他所有点连边。这样就可以把它一眼顶真出来。

发现好像只知道这一个点有点废。于是我们钦定 C 是唯一一个不与 A 连的点。然后我们又得到了一个可以被识别出来的全新点。

延续之前的思路，之前是向所有点连边，考虑让 C 只和一些点连边。比方说，让 C 与 B 连边，我们又得到了 B。

先考虑识别出链来，于是我们让 B 向原图中的所有点连边。这样不与 B 连边的点就是链上的点了。

发现无法确定链的方向。于是我们让 C 与链的端点连边。怎么区分链端点和 B ？看度数即可。

// Problem: C - (Un)labeled graphs
// URL: https://vjudge.net/contest/643391#problem/C
// Written by WRuperD

#include<bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
const int mininf = 1e9 + 7;
#define int long long
#define pb emplace_back
inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
inline void write(int x){if(x<0){x=~(x-1);putchar('-');}if(x>9)write(x/10);putchar(x%10+'0');}
#define put() putchar(' ')
#define endl puts("")
const int MAX = 3e3 + 10;

bool edge[MAX][MAX];

void work2(int n, int m){
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= n; j++){
			char ch = getchar();
			while(ch != '0' and ch != '1')	ch = getchar();
			edge[i][j] = ch - '0';
		}
	}
	for(int i = n + 1; i <= m - 3; i++){
		int now = i - n - 1;
		for(int j = 1; j <= n; j++){
			if(j & (1ll << now)){
				edge[j][i] = edge[i][j] = 1;
			}
		}
		if(i != m - 3){
			edge[i][i + 1] = edge[i + 1][i] = 1;
		}
	}
	int A = m - 2, B = m - 1, C = m;
	for(int i = 1; i < m; i++)	{
		if(i != A)	edge[i][A] = edge[A][i] = 1;
	}
	edge[B][C] = edge[C][B] = 1;
	edge[n + 1][C] = edge[C][n + 1] = 1;
	for(int i = 1; i <= n; i++){
	 	edge[B][i] = edge[i][B] = 1;
	}
	for(int i = 1; i <= m; i++){
		for(int j = 1; j <= m; j++){
			write(edge[i][j]);
		}endl;
	}
}

int id[MAX];
int deg[MAX];
int Ans[MAX][MAX];
bool isg[MAX];

void work(int x, int nowid, int m){
	for(int i = 1; i <= m; i++){
		if(edge[x][i] and isg[i]){
			id[i] |= nowid;
		}
	}
	id[x] = inf;
	isg[x] = 1;
	for(int i = 1; i <= m; i++)	if(edge[x][i] and !isg[i])	work(i, nowid * 2, m);
}

void work3(int m, int n){
	for(int i = 1; i <= m; i++){
		for(int j = 1; j <= m; j++){
			char ch = getchar();
			while(ch != '0' and ch != '1')	ch = getchar();
			edge[i][j] = ch - '0';
			deg[i] += edge[i][j], deg[j] += edge[i][j];
		}
	}
	int A = 0;
	for(int i = 1; i <= m; i++){
		if(deg[A] < deg[i])	A = i;
	}
	id[A] = n + 1;
	int C = 0;
	for(int i = 1; i <= m; i++)	if(i != A and edge[A][i] == 0) C = i;
	int B = 0;
	for(int i = 1; i <= m; i++)	if(edge[C][i] and deg[i] >= deg[B])	B = i;
	for(int i = 1; i <= m; i++)	if(edge[B][i] and i != A and i != C)	isg[i] = 1;
	id[A] = id[B] = id[C] = inf;
	isg[A] = isg[B] = isg[C] = 1;
	// write(A), put(), write(B), put(), write(C), endl;
	for(int i = 1; i <= m; i++)	if(edge[C][i] and i != B)	work(i, 1, m);
	// write(isg[2]), endl;
	for(int i = 1; i <= m; i++)	if(!id[i])	id[i] = n;
	// for(int i = 1; i <= m; i++)	write(is[])
	for(int i = 1; i <= m; i++){
		for(int j = 1; j <= m; j++){
			if(!id[i] or !id[j] or id[i] > n or id[j] > n or i == A or i == B or i == C or j == A or j == B or j == C)	continue;
			Ans[id[i]][id[j]] = edge[i][j];
			Ans[id[j]][id[i]] = edge[i][j];
		}
	}
	// for(int i = 1; i <= m; i++)	if()
	// write(A), put(), write(B), put(), write(C), endl;
	// for(int i = 1; i <= m; i++)	write(id[i]), put();
	// endl;
	// write(n), endl;
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= n; j++){
			write(Ans[i][j]);
		}endl;
	}
}

void solve(){
	int n = read(), m = read();
	if(n < m)	work2(n, m);
	else work3(n, m);
}

signed main(){
	int t = 1;
	while(t--)	solve();
	return 0;