[FJOI2015]火星商店问题

标签：ack   climits   sign   ons   double   bit   opera   查询   影响   

线段树分治。以时间轴建立线段树，每一个线段树节点，存放[L,R]时间内，有影响的操作1，建立可持久化trie树，trie树以商店位置为root，就可以支持商店的区间查询，然后将操作0，按照商店位置排序，进行线段树分治，每次到一个节点，先把操作0插入trie树，然后把所有当前时间内存的有影响的操作1全部拿出来，进行查询即可。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#include <iomanip>
#pragma GCC optimize(2)
#define up(i,a,b)  for(int i=a;i<b;i++)
#define dw(i,a,b)  for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
	char ch = getchar(); ll x = 0, f = 1;
	while (ch<‘0‘ || ch>‘9‘) { if (ch == ‘-‘)f = -1; ch = getchar(); }
	while (ch >= ‘0‘ && ch <= ‘9‘) { x = x * 10 + ch - ‘0‘; ch = getchar(); }
	return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 1e5 + 10;
struct trie {
	int ch[N * 22][2]; int root[N];
	int sum[N * 22];
	int tot;
	void clear()
	{
		tot = 0;
	}
	void insert(int &o, int pre, int x,int pos)
	{
		o = ++tot;
		sum[o] = sum[pre] + 1;
		if (pos == -1)return;
		int f = (x >> pos & 1);
		ch[o][f ^ 1] = ch[pre][f ^ 1];
		insert(ch[o][f], ch[pre][f], x, pos - 1);
	}
	int query(int l, int r, int x, int pos)
	{
		if (pos == -1)return 0;
		int f = (x >> pos & 1);
		if (sum[ch[r][f ^ 1]] - sum[ch[l][f ^ 1]])
		{
			return query(ch[l][f ^ 1], ch[r][f ^ 1], x, pos - 1) + (1 << pos);
		}
		else return query(ch[l][f], ch[r][f], x, pos - 1);
	}
}T;
int n, m;
int node_cnt, day;
struct node {
	int st, ed, l, r, x, id;
}nd[N];
struct shop {
	int d, v, p;
	bool operator<(const shop p)const {
		return d < p.d;
	}
}sp[N];
int ans[N];
struct Tree {
	vector<int> tr[N << 2];
	void update(int l, int r, int root, int L,int R,int id)
	{
		if (L > R)return;
		if (L <= l && r <= R)
		{
			tr[root].push_back(id);
			return;
		}
		int mid = (l + r) >> 1;
		if (L <= mid)update(lson, L, R, id);
		if (R > mid)update(rson,L, R, id);
	}
	void cal(int L, int R, int root)
	{
		T.clear();
		int temp = 1;
		vector<int>vec;
		upd(i, L, R)
		{
			vec.push_back(sp[i].d);
			T.insert(T.root[temp], T.root[temp - 1], sp[i].v, 20);
			temp++;
		}
		for (auto k : tr[root])
		{
			int lf = upper_bound(vec.begin(), vec.end(), nd[k].l - 1) - vec.begin();
			int rt = upper_bound(vec.begin(), vec.end(), nd[k].r) - vec.begin();
			ans[nd[k].id] = max(ans[nd[k].id], T.query(T.root[lf], T.root[rt], nd[k].x, 20));
		}
	}
	void div(int l, int r,int root, int L, int R)
	{
		if (L > R)return;
		cal(L, R, root);
		if (l == r)return;
		int mid = (l + r) >> 1;
		vector<shop>t1, t2;
		upd(i, L, R)
		{
			if (sp[i].p <= mid)t1.push_back(sp[i]);
			else t2.push_back(sp[i]);
		}
		up(i, 0, t1.size())sp[L + i] = t1[i];
		up(i, 0, t2.size())sp[L + t1.size() + i] = t2[i];
		div(lson, L, L + t1.size() - 1);
		div(rson, L + t1.size(), R);
	}
}seg;
int main()
{
	n = read(), m = read();
	int u;
	upd(i, 1, n)
	{
		u = read();
		T.insert(T.root[i], T.root[i - 1], u, 20);
	}
	int op, l, r, x, d;
	while (m--) {
		op = read();
		if (op)
		{
			l = read(), r = read(), x = read(), d = read();
			++node_cnt;
			nd[node_cnt] = node{ max(day - d + 1,1),day,l,r,x,node_cnt };
			ans[node_cnt] = T.query(T.root[l - 1], T.root[r], x, 20);
		}
		else {
			day++;
			l = read(), r = read();
			sp[day] = shop{ l,r ,day };
		}
	}
	sort(sp + 1, sp + day + 1);
	upd(i, 1, node_cnt)
		seg.update(1, day, 1, nd[i].st, nd[i].ed, nd[i].id);
	seg.div(1, day, 1, 1, day);
	upd(i, 1, node_cnt)
	{
		printf("%d\n", ans[i]);
	}
	return 0;
}

[FJOI2015]火星商店问题

标签：ack   climits   sign   ons   double   bit   opera   查询   影响   

原文地址：https://www.cnblogs.com/LORDXX/p/13268528.html