Comet OJ

大意: 给定无向图, 点$i$点权$b_i$, 边$(x,y,z)$对序列贡献是把$A[b_x oplus b_y]$加上$z$.

多组询问, 一共三种操作: 1. 修改点权. 2.修改边权. 3. 求序列$A$区间和.

图按度数分块.

对于轻点的贡献直接树状数组维护, 复杂度$O(17sqrt{m})$

每一条重边选度数较大的重点$x$建一棵$01trie$, 询问$[L,R]$区间和就等价于求异或$b[x]$后范围在$[L,R]$的和, 复杂度$O(17sqrt{m})$.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 998244353;
const int N = 131080;
const int S = 4000;
int n, m, q, x[N], y[N], z[N];
int b[N], deg[N], ID[N], E[N];
vector<int> big, small[N];
struct {
	ll c[N];
	void add(int x, int v) {
		for (++x; x<N; x+=x&-x) c[x]+=v;
	}
	ll qry(int x) {
		ll ret = 0;
		for (++x; x; x^=x&-x) ret+=c[x];
		return ret;
	}
} BIT;
struct {
	int T, tot;
	struct {int ch[2],v;} a[N<<2];
	void add(int &o, int d, int x, int v) {
		if (!o) o = ++tot;
		a[o].v = (a[o].v+v)%P;
		if (d>=0) add(a[o].ch[x>>d&1],d-1,x,v);
	}
	void add(int x, int v) {
		add(T,16,x,v);
	}
	ll qry(int o, int d, int x, int v) {
		if (d<0) return a[o].v;
		int f1 = x>>d&1, f2 = v>>d&1;
		if (f2) return a[a[o].ch[f1]].v+qry(a[o].ch[!f1],d-1,x,v);
		return qry(a[o].ch[f1],d-1,x,v);
	}
	ll qry(int x, int v) {
		ll ret = qry(T,16,x,v);
		return ret;
	}
} tr[100];
void add(int id, int tp) {
	if (E[id]) tr[E[id]].add(b[x[id]]^b[y[id]]^b[big[E[id]]],tp*z[id]);
	else BIT.add(b[x[id]]^b[y[id]],tp*z[id]);
}
int qry(int x) {
	if (x<0) return 0;
	ll ans = BIT.qry(x);
	for (int i=1; i<big.size(); ++i) {
		ans += tr[i].qry(b[big[i]],x);
	}
	return ans%P;
}

int main() {
	scanf("%d%d%d", &n, &m, &q);
	REP(i,1,n) scanf("%d", b+i);
	REP(i,1,m) {
		scanf("%d%d%d", x+i, y+i, z+i);
		++deg[x[i]],++deg[y[i]];
	}
	big.pb(0);
	REP(i,1,n) if (deg[i]>=S) { 
		ID[i] = ++*ID;
		big.pb(i);
	}
	REP(i,1,m) {
		if (ID[x[i]]&&deg[x[i]]>deg[y[i]]) {
			E[i] = ID[x[i]];
			small[y[i]].pb(i);
		}
		else if (ID[y[i]]) {
			E[i] = ID[y[i]];
			small[x[i]].pb(i);
		}
		else {
			small[x[i]].pb(i);
			small[y[i]].pb(i);
		}
		add(i, 1);
	}
	while (q--) {
		int op,x,y;
		scanf("%d%d%d", &op, &x, &y);
		if (op==1) {
			for (auto t:small[x]) add(t,-1);
			b[x] = y;
			for (auto t:small[x]) add(t,1);
		}
		else if (op==2) {
			add(x,-1);
			z[x] = y;
			add(x,1);
		}
		else { 
			int ans = (qry(y)-qry(x-1))%P;
			if (ans<0) ans += P;
			printf("%d
", ans);
		}
	}
}