DZY Loves Chinese II 题解

link

Description

给出一个无向图，每次求删掉若干条边后是否联通，强制在线。

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

Solution

挺有意思的，我们可以考虑随机化，先随便构造出一棵树，那么不连通当且仅当存在一条树边使得跨过这条边的边都被删了，那么我们可以考虑给非树边一个随即权值，然后树边的权值设为跨过它的非树边权值异或和，每次只需要判断是否存在异或和等于 \(0\) 的非空集合即可。

Code

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

#define Int register int
#define MAXN 500005

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');}
template <typename T> inline void chkmax (T &a,T b){a = max (a,b);}
template <typename T> inline void chkmin (T &a,T b){a = min (a,b);}

int n,m,fa[MAXN];
int findSet (int x){return fa[x] == x ? x : fa[x] = findSet (fa[x]);}

struct edge{
	int u,v,val;
}e[MAXN];
bool ing[MAXN];

#define pii pair<int,int>
#define se second
#define fi first
vector <pii> g[MAXN];

int par[MAXN],dep[MAXN],det[MAXN];
void dfs (int u,int fa){
	par[u] = fa;
	for (pii it : g[u]) if (it.fi ^ fa) dfs (it.fi,u);
}

void dfs1 (int u,int fa){
	for (pii it : g[u]) if (it.fi ^ fa) dfs1 (it.fi,u),e[it.se].val = det[it.fi],det[u] ^= det[it.fi];
}

int d[31];
void ins (int &v){
	for (Int i = 30;~i;-- i) if (v >> i & 1){
		if (d[i]) v ^= d[i];
		else{
			d[i] = v;
			break;
		}
	}
}

signed main(){
	srand (time(NULL)),read (n,m);
	for (Int u = 1;u <= n;++ u) fa[u] = u;
	for (Int i = 1;i <= m;++ i){
		read (e[i].u,e[i].v);
		if (findSet (e[i].u) == findSet (e[i].v)) e[i].val = rand();
		else
			fa[fa[e[i].u]] = fa[e[i].v],
			g[e[i].u].push_back ({e[i].v,i}),
			g[e[i].v].push_back ({e[i].u,i});
	}
	dfs (1,0);
	for (Int i = 1;i <= m;++ i) if (e[i].val) det[e[i].u] ^= e[i].val,det[e[i].v] ^= e[i].val;
	dfs1 (1,0);
	int q,cnt = 0;read (q);
	while (q --> 0){
		int K;read (K),memset (d,0,sizeof (d));bool flg = 0;
		for (Int i = 1,v;i <= K;++ i) read (v),v ^= cnt,v = e[v].val,ins (v),flg |= (v == 0);
		cnt += (!flg),puts (flg ? "Disconnected" : "Connected");
	}
	return 0;
}