AT_abc334_g [ABC334G] Christmas Color Grid 2 题解

对于这个网格图四联通建边，然后建圆方树。

很显然一个图中删掉一个点增加的连通块数量为圆方树上这个点的度数 $-1$。

于是直接就做完了。每个点累加答案，求个平均数就好了。

#include <bits/stdc++.h>
using namespace std;
#define int long long

const int N = 1005, MOD = 998244353, HSMOD = 1610612741, HSMOD2 = 998244353; // Remember to change

int n, m, q, t, a[N];
char c[N][N];

namespace FastIo
{
	#define QUICKCIN ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
	int read()
	{
		char ch = getchar();
		int x = 0, f = 1;
		while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
		while (ch == '-')
		{
			f = -f;
			ch = getchar();
		}
		while (ch >= '0' && ch <= '9')
		{
			x = (x << 1) + (x << 3) + (ch ^ 48);
			ch = getchar();
		}
		return x * f;
	}
	template<class T>
	void write(T x)
	{
		if (x < 0)
		{
			putchar('-');
			x = -x;
		}
		if (x > 9) write(x / 10);
		putchar(x % 10 + '0');
	}
	template<class T>
	void writeln(T x)
	{
		write(x);
		putchar('\n');
	}
}

int qpow(int a, int b)
{
	int res=1LL,base=a;
	while(b)
	{
		if(b&1LL)
		{
			res=res*base%MOD;
		}
		base=base*base%MOD;
		b>>=1LL;
	}
	return res;
}

int get(int x, int y)
{
	return (x-1)*m+y;
}

class Union_Find
{
public:
	int fa[N*N],sz[N*N];
	void Init()
	{
		for(int i=1;i<N*N;i++) fa[i]=i,sz[i]=1;
	}
	int find(int u)
	{
		return (fa[u]==u?u:find(fa[u])); 
	}
	pair<int, int> merge(int u, int v)
	{
		if((u=find(u))==(v=find(v))) return make_pair(-1,-1);
		if(sz[u]<sz[v]) swap(u,v);
		sz[u]+=sz[v];
		fa[v]=u;
		return make_pair(u,v);
	}
	void del(int u, int v)
	{
		sz[u]-=sz[v];
		fa[v]=v;
	}
}uf;

bool vis[N][N];
int dx[]={0,0,-1,1};
int dy[]={-1,1,0,0};

void dfs(int x, int y)
{
	vis[x][y]=1;
	for(int i=0;i<4;i++)
	{
		int nx=x+dx[i],ny=y+dy[i];
		if(nx>=1&&nx<=n&&ny>=1&&ny<=m&&!vis[nx][ny]&&c[nx][ny]=='#')
		{
			dfs(nx,ny);
			uf.merge(get(x,y),get(nx,ny));
		}
	}
}

int dfn[5*N*N],low[5*N*N];
vector<int> NG[5*N*N];
int idx;
int stk[5*N*N],top;
vector<int> G[5*N*N];
int CC;

void tarjan(int u)
{
	dfn[u]=low[u]=++idx;
	stk[++top]=u;
	for(auto&j:G[u])
	{
		if(!dfn[j])
		{
			tarjan(j);
			low[u]=min(low[u],low[j]);
			if(low[j]==dfn[u])
			{
				CC++;
				for(int y=0;y^j;top--)
				{
					y=stk[top];
					NG[CC].emplace_back(y);
					NG[y].emplace_back(CC);
				}
				NG[u].emplace_back(CC);
				NG[CC].emplace_back(u);
			}
		}
		else low[u]=min(low[u],dfn[j]);
	}
}

signed main()
{
	ios::sync_with_stdio(0), cin.tie(nullptr), cout.tie(nullptr);
	cin >> n >> m;
	CC=n*m;
	uf.Init();
	int cnt=0;
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			cin>>c[i][j];
			cnt+=(c[i][j]=='#');
		}
	}
	int cc=0;
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			for(int k=0;k<4;k++)
			{
				int ni=i+dx[k],nj=j+dy[k];
				if(ni>=1&&ni<=n&&nj>=1&&nj<=m&&c[i][j]=='#'&&c[ni][nj]=='#')
				{
					G[get(i,j)].emplace_back(get(ni,nj));
				}
			}
			if(c[i][j]=='#'&&!vis[i][j]) 
			{
				dfs(i,j),cc++;
			}
		}
	}
	int ans=0;
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			if(c[i][j]=='#')
			{
				if(!dfn[get(i,j)])
				{
					top=0;
					tarjan(get(i,j));
				}
			}
		}
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			if(c[i][j]=='#')
			{
				if(!dfn[get(i,j)])
				{
					top=0;
					tarjan(get(i,j));
				}
			}
		}
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			if(c[i][j]=='#')
			{
				int nc=cc+NG[get(i,j)].size()-1;
				ans=(ans+nc*qpow(cnt,MOD-2LL)%MOD)%MOD; 
			}
		}
	}
	cout<<ans<<"\n";
	return 0;
}