2020 CCPC Wannafly Winter Camp Day1 I- K小数查询

分块+暴力。
对于每一块来说。如果更新不是一整块,我们直接暴力重构,如果是一整块更新的话,就给这一个块打上一个标记。
当询问的时候,我们二分答案,对于不是一整块的区间,暴力。
对于是一整块的区间,我们优先判断他和标记的大小。如果标记更大,那么由于这一道题更新的时候,区间只取min,故块中的所有数据不会因为更新而变的更大,因此,我们直接在块里面进行二分得答案。(如果这个块不是一整块更新,那么一定被重构过)。

#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>
#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 = 8e4 + 10;
int n, m, base;
int belong[N];
int lf[N], rt[N];
int a[N];
int mx[N];
vector<int>vec[N];
void build()
{
	int sz = belong[n];
	upd(i, 1, sz)
	{
		upd(j, lf[i], rt[i])
		{
			vec[i].push_back(a[j]);
		}
		sort(vec[i].begin(), vec[i].end());
		mx[i] = vec[i].back();
	}
}
void rebuild(int pos)
{
	vec[pos].clear();
	upd(i, lf[pos], rt[pos])
	{
		a[i] = min(a[i], mx[pos]);
		vec[pos].push_back(a[i]);
	}
	sort(vec[pos].begin(), vec[pos].end());
	mx[pos] = vec[pos].back();
}
void update(int l,int r,int val)
{
	int x = belong[l]; int y = belong[r];
	if (x == y)
	{
		upd(i, l, r)
		{
			a[i] = min(a[i],val);
		}
		rebuild(x);
		return;
	}
	upd(i, l, rt[x])
	{
		a[i] = min(a[i], val);
	}
	rebuild(x);
	upd(i, lf[y], r)
	{
		a[i] = min(a[i], val);
	}
	rebuild(y);
	up(i, x + 1, y)
		mx[i] = min(val, mx[i]);
}
int query(int l, int r, int val)
{
	int x = belong[l]; int y = belong[r];
	int res = 0;
	if (x == y)
	{
		upd(i, l, r)
		{
			if (a[i] <= val)
				res++;
		}
		return res;
	}
	upd(i, l, rt[x]) {
		if (a[i] <= val)res++;
	}
	upd(i, lf[y], r) {
		if (a[i] <= val)res++;
	}
	up(i,x+1,y)
	{
		if (mx[i] <= val) {
			res += rt[i] - lf[i] + 1;
		}
		else
		{
			int temp = upper_bound(vec[i].begin(), vec[i].end(), val)-vec[i].begin();
			res += temp;
		}
	}
	return res;
}
int judge(int l,int r,int k)
{
	int tl = 0, tr = 1e9 + 1;
	int x = belong[l]; int y = belong[r];
	rebuild(x);
	if (x != y)rebuild(y);
	while (tr - 1 > tl)
	{
		int mid = (tr + tl) >> 1;
		if (query(l, r, mid) >= k)tr = mid;
		else tl = mid;
	}
	return tr;
}
int main()
{
	n = read(); m = read();
	upd(i, 1, n)a[i] = read();
	base = sqrt(n) + 1;
	memset(lf, INF, sizeof(lf));
	memset(rt, 0, sizeof(rt));
	upd(i, 1, n)
	{
		int pos= i / base + 1;
		belong[i] = pos;
		lf[pos] = min(lf[pos], i);
		rt[pos] = max(rt[pos], i);
	}
	build();
	int templ, tempr, tempval;
	int op;
	while (m--)
	{
		op = read();
		if (op == 1)
		{
			templ = read(); tempr = read(); tempval = read();
			update(templ, tempr, tempval);
		}
		else
		{
			templ = read(); tempr = read(); tempval = read();
			printf("%d\n", judge(templ, tempr, tempval));
		}
	}
	return 0;
}
posted @ 2020-03-06 21:25  LORDXX  阅读(96)  评论(0编辑  收藏  举报