【BZOJ】3123: [Sdoi2013]森林

题解

………………………………………………

我莫不是一个智障吧

我把testdata的编号

当成数据组数读进来

我简直有毒

以为哪里写错了自闭了好久

实际上这题很简单，只要愉悦地开个启发式合并，然后每次暴力修改一个点的根缀主席树和倍增lca数组就行

复杂度\(n \log^2 n\)

代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 80005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
	res = 0;T f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		res = res * 10 + c - '0';
		c = getchar();
	}
	res *= f;
}
template<class T>
void out(T x) {
	if(x < 0) {x = -x;putchar('-');}
	if(x >= 10) {
		out(x / 10);
	}
	putchar('0' + x % 10);
}
int N,M,Q;
int val[MAXN],w[MAXN],cnt,fa[MAXN][20],belong[MAXN],siz[MAXN],dep[MAXN];
struct node {
	int sum,lc,rc;
}tr[MAXN * 200];
struct Enode {
	int to,next;
}E[MAXN * 2];
int Ncnt,rt[MAXN],sumE,head[MAXN];
void add(int u,int v) {
	E[++sumE].to = v;
	E[sumE].next = head[u];
	head[u] = sumE;
}
int getfa(int u) {
	return belong[u] == u ? u : belong[u] = getfa(belong[u]);
}
void Insert(const int &x,int &y,int l,int r,int v) {
	y = ++Ncnt;
	tr[y] = tr[x];
	++tr[y].sum;
	if(l == r) return;
	int mid = (l + r) >> 1;
	if(v <= mid) Insert(tr[x].lc,tr[y].lc,l,mid,v);
	else Insert(tr[x].rc,tr[y].rc,mid + 1,r,v);
}
void rebuild(int u,int f) {
	dep[u] = dep[f] + 1;
	Insert(rt[f],rt[u],1,cnt,w[u]);
	fa[u][0] = f;
	for(int i = 1 ; i <= 18 ; ++i) {
		fa[u][i] = fa[fa[u][i - 1]][i - 1];
	}
	for(int i = head[u] ; i ; i = E[i].next) {
		int v = E[i].to;
		if(v != f) rebuild(v,u);
	}
}
void Init() {
	Ncnt = 0;sumE = 0;
	memset(fa,0,sizeof(fa));
	memset(rt,0,sizeof(rt));
	memset(head,0,sizeof(head));
	memset(dep,0,sizeof(dep));
	read(N);read(M);read(Q);
	for(int i = 1 ; i <= N ; ++i) {
		read(w[i]);val[i] = w[i];
	}
	sort(val + 1,val + N + 1);
	cnt = unique(val + 1,val + N + 1) - val - 1;
	for(int i = 1 ; i <= N ; ++i) {
		w[i] = lower_bound(val + 1,val + cnt + 1,w[i]) - val;
	}
	for(int i = 1 ; i <= N ; ++i) belong[i] = i,siz[i] = 1;
	int u,v;
	for(int i = 1 ; i <= M ; ++i) {
		read(u);read(v);
		add(u,v);add(v,u);
		siz[getfa(v)] += siz[getfa(u)];
		belong[getfa(u)] = getfa(v);
	}
	for(int i = 1 ; i <= N ; ++i) {
		if(i == belong[i]) rebuild(i,0);
	}
}
int lca(int u,int v) {
	if(dep[u] < dep[v]) swap(u,v);
	int l = 18;
	while(dep[u] > dep[v]) {
		if(dep[fa[u][l]] >= dep[v]) u = fa[u][l];
		--l;
	}
	if(u == v) return u;
	l = 18;
	while(fa[u][0] != fa[v][0]) {
		if(fa[u][l] != fa[v][l]) {
			u = fa[u][l];
			v = fa[v][l];
		}
		--l;
	}
	return fa[u][0];
}
int Query(int u,int v,int f,int k) {
	int L = 1,R = cnt;
	int t = w[f];
	u = rt[u],v = rt[v],f = rt[f];
	while(L < R) {
		int mid = (L + R) >> 1;
		int s = tr[tr[u].lc].sum - tr[tr[f].lc].sum + tr[tr[v].lc].sum - tr[tr[f].lc].sum;
		if(t >= L && t <= mid) ++s;
		if(s < k) {
			k -= s;
			L = mid + 1;
			u = tr[u].rc;v = tr[v].rc;f = tr[f].rc;
		}
		else {
			R = mid;
			u = tr[u].lc;v = tr[v].lc;f = tr[f].lc;
		}
	}
	return L;
}
void Solve() {
	char op[5];int x,y,k;
	int lastans = 0;
	for(int i = 1 ; i <= Q ; ++i) {
		scanf("%s",op + 1);
		read(x);read(y);
		x ^= lastans;y ^= lastans;
		if(op[1] == 'L') {
			int p = getfa(x),q = getfa(y);
			if(siz[p] > siz[q]) {swap(x,y);swap(p,q);}
			belong[p] = q;siz[q] += siz[p];
			rebuild(x,y);
			add(x,y);add(y,x);
		}
		else {
			read(k);
			k ^= lastans;
			out(lastans = val[Query(x,y,lca(x,y),k)]);enter;
		}
		//out(i);enter;
	}
}
int main() {
#ifdef ivorysi
	freopen("f1.in","r",stdin);
#endif
	int T;
	read(T);
	Init();
	Solve();
}