[ARC165D] Substring Comparison

Problem Statement

For an integer sequence $X=(X_1,X_2,\dots,X_n)$, let $X[L,R]$ denote the integer sequence $(X_L,X_{L+1},\dots,X_{R})$.

You are given integers $N$ and $M$, and $M$ quadruples of integers $(A_i,B_i,C_i,D_i)$.

Determine if there is an integer sequence $X$ of length $N$ that satisfies the following condition for every $i=1,2,\dots,M$:

• $X[A_i,B_i]$ is lexicographically smaller than $X[C_i,D_i]$.

What is lexicographical order on sequences?

A sequence $S = (S_1,S_2,\ldots,S_{|S|})$ is lexicographically smaller than $T = (T_1,T_2,\ldots,T_{|T|})$ when 1. or 2. below holds. Here, $|S|$ and $|T|$ denotes the lengths of $S$ and $T$, respectively.

1. $|S| \lt |T|$ and $(S_1,S_2,\ldots,S_{|S|}) = (T_1,T_2,\ldots,T_{|S|})$.
2. There is an integer $1 \leq i \leq \min\lbrace |S|, |T| \rbrace$ that satisfy both of the following:
   • $(S_1,S_2,\ldots,S_{i-1}) = (T_1,T_2,\ldots,T_{i-1})$.
   • $S_i$ is smaller than $T_i$ (as a number).

Constraints

• $2 \leq N \leq 2000$
• $1 \leq M \leq 2000$
• $1 \leq A_i \leq B_i \leq N$
• $1 \leq C_i \leq D_i \leq N$
• All input values are integers.

首先一定满足 $X_{A_i}\le X_{C_i}$，从 $A_i$ 向 $C_i$ 连边。
跑一次 tarjan 后，此时如果出现了大于 1 的强连通分量，那么这个分量里所有的数都是一样的，用一个并查集并起来。
如果这个强连通分量里的边有 $i$，那么说明 $X_{A_{i+1}}\le X_{C_{i+1}}$
以此类推，知道某一次没有大于 1 的强连通分量，就结束了。
发现每次至少合并两个点，最多跑 $N$ 次 tarjan。同时每次图中至多有 $M$ 条边，所以复杂度是 $O(NM)$ 的。

#include<bits/stdc++.h>
using namespace std;
const int N=2005;
int n,m,k,tp,st[N],hd[N],fl,e_num,tme,p[N],id[N],dfn[N],low[N],idx,a[N],b[N],c[N],d[N],fa[N];
struct edge{
	int v,nxt;
}e[N];
void add_edge(int u,int v)
{
	e[++e_num]=(edge){v,hd[u]};
	hd[u]=e_num;
}
int find(int x)
{
	if(fa[x]==x)
		return x;
	return fa[x]=find(fa[x]);
}
void tarjan(int x)
{
	dfn[x]=low[x]=++tme;
	st[++tp]=x;
	for(int i=hd[x];i;i=e[i].nxt)
	{
		if(!dfn[e[i].v])
			tarjan(e[i].v),low[x]=min(low[x],low[e[i].v]);
		else if(!id[e[i].v])
			low[x]=min(low[x],dfn[e[i].v]);
	}
	if(low[x]==dfn[x])
	{
		++idx;
		if(st[tp]^x)
		{
			fl=1;
			while(st[tp]^x)
				id[st[tp]]=idx,fa[find(st[tp--])]=find(x);
		}
		id[st[tp--]]=idx;
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		fa[i]=i;
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d%d%d",a+i,b+i,c+i,d+i);
		add_edge(a[i],c[i]);
	}
	while(1)
	{
		fl=tme=idx=0;
		memset(dfn,0,sizeof(dfn));
		memset(id,0,sizeof(id));
		for(int i=1;i<=n;i++)
			if(!dfn[i])
				tarjan(i);
		if(!fl)
			break;
		memset(hd,e_num=0,sizeof(hd));
		for(int i=1;i<=m;i++)
		{
			while(a[i]+p[i]<=b[i]&&c[i]+p[i]<=d[i]&&find(a[i]+p[i])==find(c[i]+p[i]))
				++p[i];
			if(a[i]+p[i]>b[i])
			{
				if(d[i]-c[i]<=b[i]-a[i])
					return puts("No"),0;
			}
			else if(c[i]+p[i]>d[i])
				return puts("No"),0;
			else
				add_edge(find(a[i]+p[i]),find(c[i]+p[i]));
		}
	}
	puts("Yes");
}