luogu 3295[SCOI2016]萌萌哒

题目链接：luogu 3295

很明显的想法是用并查集维护最后有几个块内的数字必须相等，设最后有(cnt)个联通块，那么答案就是(9 imes10^{cnt-1})

暴力维护并查集是(O(nlen))的，显然超时，考虑优化

记(f_{i,j})表示区间([i,i+2^j-1])的编号，编号相同则区间中元素值相同

考虑将一段区间((l_1,r_1))和((l_2,r_2))合并时，将每个区间拆成(log)段，合并((l_1,l_1+2^k-1))和((l_2,l_2+2^k-1))

最后像线段树标记下放一样将(f_{i,j})的值下放到(f_{i+2^j-1,j-1})和(f_{i+2^j,j-1})

刚开始一只啊在想线段树优化然后。。。就凉了

#include<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<vector>
#include<queue>
#include<map>
#include<set>
using namespace std;
#define lowbit(x) (x)&(-x)
#define rep(i,a,b) for (int i=a;i<=b;i++)
#define per(i,a,b) for (int i=a;i>=b;i--)
#define maxd 1000000007
typedef long long ll;
const int N=100000;
const double pi=acos(-1.0);
int n,m,fa[500100][20];

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*10+(ch-'0');ch=getchar();}
	return x*f;
}

int find(int x,int y)
{
	if (fa[x][y]==x) return x;
	fa[x][y]=find(fa[x][y],y);
	return fa[x][y];
}

int main()
{
	n=read();m=read();
	rep(i,1,n)
		rep(j,0,19) fa[i][j]=i;
	rep(i,1,m)
	{
		int l1=read(),r1=read(),l2=read(),r2=read();
		int pos1=l1,pos2=l2;
		per(j,19,0)
		{
			if (pos1+(1<<j)-1<=r1)
			{
				int fa1=find(pos1,j),fa2=find(pos2,j);
				if (fa1!=fa2) fa[fa1][j]=fa2;
				pos1+=(1<<j);pos2+=(1<<j);
			}
		}
	}
	per(j,19,1)
	{
		int i;
		for (i=1;i+(1<<j)-1<=n;i++)
		{
			int fa1=find(i,j),fa2=find(i,j-1);
			fa[fa2][j-1]=find(fa1,j-1);
			int fa3=find(i+(1<<(j-1)),j-1);
			fa[fa3][j-1]=find(fa1+(1<<(j-1)),j-1);
		}
	}
	ll ans=1;
	rep(i,1,n)
	{
		if (fa[i][0]==i) 
		{
			if (ans==1) ans=(ans*9)%maxd;
			else ans=(ans*10)%maxd;
		}
	}
	printf("%lld",ans);
	return 0;
}