[BZOJ4569][SCOI2016]萌萌哒

bzoj
luogu

sol

维护\(\log{n}\)个并查集，每个限制就行\(ST\)表那样分成两个区间然后合并。
接着把所有合并标记下方，最后就只要查第\(0\)个并查集里有几个代表元就好了。

code

#include<cstdio>
#include<algorithm>
using namespace std;
int gi()
{
	int x=0,w=1;char ch=getchar();
	while ((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
	if (ch=='-') w=0,ch=getchar();
	while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
	return w?x:-x;
}
const int N = 2e5+5;
const int mod = 1e9+7;
int n,m,lg[N],ans;
struct bcj{
	int fa[N];
	void init(){for (int i=1;i<=n;++i) fa[i]=i;}
	int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
	void merge(int i,int j){fa[find(i)]=find(j);}
}S[18];
int fastpow(int a,int b)
{
	int res=1;
	while (b) {if (b&1) res=1ll*res*a%mod;a=1ll*a*a%mod;b>>=1;}
	return res;
}
int main()
{
	n=gi();m=gi();
	for (int i=2;i<=n;++i) lg[i]=lg[i>>1]+1;
	for (int i=0;i<=lg[n];++i) S[i].init();
	for (int i=1;i<=m;++i)
	{
		int a=gi(),b=gi(),c=gi(),d=gi(),k=lg[b-a+1];
		S[k].merge(a,c);S[k].merge(b-(1<<k)+1,d-(1<<k)+1);
	}
	for (int j=lg[n];j;--j)
		for (int i=1;i<=n;++i)
		{
			int x=S[j].find(i);
			S[j-1].merge(i,x);S[j-1].merge(i+(1<<j-1),x+(1<<j-1));
		}
	for (int i=1;i<=n;++i) if (S[0].find(i)==i) ++ans;
	printf("%lld\n",9ll*fastpow(10,ans-1)%mod);return 0;
}