[JZOJ 5782] 城市猎人

思路：

并查集按秩合并维护出现时间。

最早连接时间就是树上连接最大值。

\(qwq\)我居然把路径压缩和按秩合并打到一个程序里了...OvO

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000010;
struct edge {
	int to;
	int nxt;
	int w;
}e[maxn << 1];
//struct asks{
//	int x,y;
//}q[maxn<<1];
int n,m,q,cnt;
int x,y;
int rank[maxn];
int fa[maxn];
int head[maxn];
inline int find(int x) {
	return x == fa[x] ? x : find(fa[x]);
} 
inline void Add_edge(int u,int v,int w) {
	e[++cnt].w = w;
	if(rank[u] > rank[v]) {
		fa[v] = u;
		e[cnt].to = u;
		e[cnt].nxt = head[v];
		head[v] = cnt;
	}
	else {
		fa[u] = v;
		e[cnt].to = v;
		e[cnt].nxt = head[u];
		head[u] = cnt;
		if(rank[u] == rank[v]) rank[u] ++;
	}
	return;
}
inline int query(int x,int y) {
	int dx = 0;
	int dy = 0;
	int res = 0;
	int l = x;
	int r = y;
	while(fa[l] != l) {
		l = fa[l];
		dx++;
	}
	while(fa[r] != r) {
		r = fa[r];
		dy ++;
	}
	if(dx < dy) {
		swap(dx,dy);
		swap(x,y);
	}
	while(dx > dy) {
		res = max(e[head[x]].w,res);
		x = fa[x];
		dx --;
	}
	if(x == y) return res;
	while(x != y) {
		res = max(res,max(e[head[x]].w,e[head[y]].w));
		x = fa[x];y = fa[y];
	}
	return res;
}
int main () {
	#ifdef ONLINE_JUDGE
		freopen("pictionary.in","r",stdin);
		freopen("pictionary.out","w",stdout);
	#endif
	scanf("%d %d %d",&n,&m,&q);
	for(int i = 1;i <= n; ++i) {
		fa[i] = i;
	}
	for(int i = 1;i <= m; ++i){
		int d = m - i + 1;
		for(int j = d*2;j <= n;j += d) {
			Add_edge(find(d),find(j),i);
			//cout<<d << ' '<< j<<endl;
		}
	}
	for(int i = 1;i <= q; ++i) {
		scanf("%d %d",&x,&y);
		printf("%d\n",query(x,y));
	}
	return 0;
}