题解 CF555E Case of Computer Network

题目传送门

题目大意

给出一个\(n\)个点\(m\)条边的无向图，有\(q\)次有向点对\((s,t)\)，问是否存在一种方法定向每条边使得每个点对可以\(s\to t\)。

\(n,m,q\le 2\times 10^5\)

思路

首先我们可以发现，一个边双连通分量里面肯定可以满足，因为任意两点之间都有两条及以上路径，于是可以一条过去，一条回来。于是，我们就可以先双连通分量缩点一下。

接着我们发现缩点完之后一定是个森林，因为如果存在环的话一定还可以缩点。那我们就可以用树上差分解决这个问题了。具体的话对于点对\((s,t)\)，显然\(s\to lca(s,t)\)应该是向上的，\(lca(s,t)\to t\)应该是向下的。如果存在一个边既向上又向下那显然是矛盾的。

于是，我们就可以在\(\Theta(n\log n)\)的时间复杂度内解决这个问题。（求lca）

\(\texttt{Code}\)

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

#define Int register int
#define MAXN 400005 

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}

bool vis[MAXN << 1];
int n,m,q,s[MAXN],t[MAXN],sum[MAXN],dfn[MAXN],low[MAXN];
struct SolveTree{
	int up;//表示边双连通分量有多少个 
	struct edge{
		int v,nxt;
	}e[MAXN << 1];
	int toop = 1,Index,s1[MAXN],s2[MAXN],fa[MAXN][21],col[MAXN],dep[MAXN],head[MAXN];
	void Add_Edge (int u,int v){
		e[++ toop] = edge {v,head[u]};head[u] = toop;
		e[++ toop] = edge {u,head[v]};head[v] = toop;
	}	
	void dfs (int u,int par){
		col[u] = Index,fa[u][0] = par,dep[u] = dep[par] + 1;
		for (Int i = 1;i <= 20;++ i) fa[u][i] = fa[fa[u][i - 1]][i - 1];
		for (Int i = head[u];i;i = e[i].nxt){
			int v = e[i].v;
			if (v ^ par) dfs (v,u);
		}
	}
	int LCA (int x,int y){
		if (dep[x] < dep[y]) swap (x,y);
		for (Int i = 20,dis = dep[x] - dep[y];~i;-- i) if (dis >> i & 1) x = fa[x][i];
		if (x == y) return x;
		else{
			for (Int i = 20;~i;-- i) if (fa[x][i] ^ fa[y][i]) x = fa[x][i],y = fa[y][i];
			return fa[x][0];
		} 
	}
	bool Remain(int u){
		for (Int i = head[u];i;i = e[i].nxt){
			int v = e[i].v;
			if (v ^ fa[u][0]){
				if (!Remain (v) || (s1[v] && s2[v])) return 0;
				s1[u] += s1[v],s2[u] += s2[v];
			}
		}
		return 1;
	}
	void Solve (){
		for (Int i = 1;i <= up;++ i) if (!col[i]) ++ Index,dfs (i,0);
		for (Int i = 1;i <= q;++ i){
			int u = s[i],v = t[i];
			if (col[u] != col[v]) return puts ("No"),void ();
			else{
				int w = LCA (u,v);
				s1[u] ++,s1[w] --,s2[v] ++,s2[w] --;
			}
		}
		for (Int i = 1;i <= up;++ i) if (dep[i] == 1 && !Remain (i)) return puts ("No"),void ();
		puts ("Yes");
	}
}T;
struct CutTree{
	struct edge{
		int v,nxt;
	}e[MAXN << 1];
	int top = 1,Index,head[MAXN],col[MAXN];
	void Add_Edge (int u,int v){
		e[++ top] = edge {v,head[u]};head[u] = top;
		e[++ top] = edge {u,head[v]};head[v] = top;
	}
	int id;
	void Tarjan (int u,int fa){
		dfn[u] = low[u] = ++ id;
		for (Int i = head[u];i;i = e[i].nxt) if (e[i].v ^ fa){
			int v = e[i].v;
			if (!dfn[v]){
				Tarjan (v,u),low[u] = min (low[u],low[v]);
				if (low[v] > dfn[u]) vis[i] = vis[i ^ 1] = 1;//vis[i]表示该边是不是割边 
			}
			else low[u] = min (low[u],dfn[v]);
		}
	}
	void dfs (int u){
		col[u] = Index;
		for (Int i = head[u],v;i;i = e[i].nxt) if (!col[v = e[i].v] && !vis[i]) dfs (v);
	}
	void build (){
		read (n,m,q);
		for (Int i = 1,u,v;i <= m;++ i) read (u,v),Add_Edge (u,v);
		for (Int i = 1;i <= n;++ i) if (!dfn[i]) Tarjan (i,0);
		for (Int i = 1;i <= n;++ i) if (!col[i]) ++ Index,dfs (i);T.up = Index;
		for (Int i = 1;i <= n;++ i) for (Int j = head[i],v;j;j = e[j].nxt) if (col[v = e[j].v] > col[i]) T.Add_Edge (col[i],col[v]); 
		for (Int i = 1;i <= q;++ i){
		 read (s[i],t[i]),s[i] = col[s[i]],t[i] = col[t[i]];
		 if (s[i] == t[i]) i --,q --; 
		 } 
	}
}G;

signed main(){ 
	G.build(),T.Solve();
	return 0;
}