8.13 杂题

P1471 方差

我们把方差公式展开:

所以只需要维护一个区间平方和和区间和

当我们更新一个区间加时

\(pushdown\) 易得

注意 \(double\) 的使用

#include <bits/stdc++.h>
using namespace std;
#define mid (l+r>>1)
#define endl '\n'
#define inl inline
#define eb emplace_back
#define ls p<<1
#define rs p<<1|1
#define lson ls,l,mid
#define rson rs,mid+1,r
const int N = 1e5 + 5;
// char buf[1<<24] , *p1 , *p2;
// #define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<24,stdin),p1==p2)?EOF:*p1++)
#define getchar() cin.get();
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f ;
}

int n , m;

double a[N];

struct Tree
{
	struct node { double sum , sqrsum , add; } t[N<<2];
	inl void pushadd ( int p , int l , int r , double val ) { t[p].add += val , t[p].sqrsum += 2 * t[p].sum * val + ( r - l + 1 ) * val * val , t[p].sum += ( r - l + 1 ) * val; }
	inl void down ( int p , int l , int r ) { if ( t[p].add ) pushadd ( lson , t[p].add ) , pushadd ( rson , t[p].add ) , t[p].add = 0; }
	inl void up ( int p ) { t[p].sum = t[ls].sum + t[rs].sum , t[p].sqrsum = t[ls].sqrsum + t[rs].sqrsum; }
	void build ( int p , int l , int r )
	{
		if ( l == r ) return t[p].sum = a[l] , t[p].sqrsum = a[l] * a[l] , void();
		build ( lson ) , build ( rson ) , up(p);
	}
	void upd ( int p , int l , int r , int x , int y , double val )
	{
		if ( x <= l && r <= y ) return pushadd ( p , l , r , val ) , void();
		down ( p , l , r );
		if ( x <= mid ) upd ( lson , x , y , val );
		if ( mid + 1 <= y ) upd ( rson , x , y , val );
		up(p);
	}
	double query1 ( int p , int l , int r , int x , int y )
	{	
		if ( x <= l && r <= y ) return t[p].sum;
		double res = 0; down ( p , l , r );
		if ( x <= mid ) res += query1 ( lson , x , y );
		if ( mid + 1 <= y ) res += query1 ( rson , x , y );
		return res;
	}
	double query2 ( int p , int l , int r , int x , int y )
	{
		if ( x <= l && r <= y ) return t[p].sqrsum;
		double res = 0; down ( p , l , r );
		if ( x <= mid ) res += query2 ( lson , x , y );
		if ( mid + 1 <= y ) res += query2 ( rson , x , y );
		return res;
	}
}T;

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	n = read() , m = read();
	for ( int i = 1 ; i <= n ; i ++ ) cin >> a[i];
	T.build ( 1 , 1 , n );
	for ( int i = 1 ; i <= m ; i ++ )
	{
		int op = read() , l = read() , r = read(); double w;
		if ( op == 1 ) cin >> w , T.upd ( 1 , 1 , n , l , r , w );
		if ( op == 2 ) cout << fixed << setprecision(4) << T.query1 ( 1 , 1 , n , l , r ) / ( 1.0 * ( r - l + 1 ) ) << endl;
		if ( op == 3 )
		{
			double temp1 = T.query1 ( 1 , 1 , n , l , r ) / ( 1.0 * ( r - l + 1 ) );
			double temp2 = T.query2 ( 1 , 1 , n , l , r ) / ( 1.0 * ( r - l + 1 ) );
			cout << fixed << setprecision(4) << temp2 - temp1 * temp1 << endl;
		}
	}
	return 0;
}

P5906 【模板】回滚莫队&不删除莫队

回滚莫队总是细节很多......

这题加点操作很好维护但是删点操作很难维护那么考虑使用回滚莫队

那么我们记录\(st[a[i]]\)数组和\(ed[a[i]]\)数组表示最早的\(a[i]\)的出现位置即可

对于滚动的左端点单独记录\(tempst\)和\(temped\)两个数组即可

#include <bits/stdc++.h>
using namespace std;
#define mid (l+r>>1)
#define endl '\n'
#define inl inline
#define eb emplace_back
const int N = 2e5 + 5;
char buf[1<<24] , *p1 , *p2;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<24,stdin),p1==p2)?EOF:*p1++)
// #define getchar() cin.get();
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f ;
}

int n , m , a[N] , st[N] , ed[N] , cnt[N] , bel[N] , block , tot , ll[N] , rr[N] , tmp[N] , l , r , lstblock , ans[N] , res , tempst[N] , temped[N];

vector<int>lsh;

struct que { int l , r , id; } q[N];


signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);

	n = read();
	for ( int i = 1 ; i <= n ; i ++ ) a[i] = read() , lsh.eb(a[i]);
	sort ( lsh.begin() , lsh.end() );
	lsh.erase ( unique ( lsh.begin() , lsh.end() ) , lsh.end() );
	for ( int i = 1 ; i <= n ; i ++ ) a[i] = lower_bound ( lsh.begin() , lsh.end() , a[i] ) - lsh.begin();

	block = sqrt(n) , tot = n / block;
	if ( n % block ) tot ++;
	for ( int i = 1 ; i <= n ; i ++ ) bel[i] = ( i - 1 ) / block + 1;
	for ( int i = 1 ; i <= tot ; i ++ ) ll[i] = ( i - 1 ) * block + 1 , rr[i] = i * block; rr[tot] = n;

	m = read();
	for ( int i = 1 ; i <= m ; i ++ ) q[i].l = read() , q[i].r = read() , q[i].id = i;
	sort ( q + 1 , q + m + 1 , [](const que &a , const que &b) { return bel[a.l] == bel[b.l] ? a.r < b.r : bel[a.l] < bel[b.l]; } );
	l = 1 , r = 0;	
	for ( int i = 1 ; i <= m ; i ++ )
	{
		if ( bel[q[i].l] == bel[q[i].r] )
		{
			int temp = 0;
			for ( int j = q[i].l ; j <= q[i].r ; j ++ ) tmp[a[j]] = j;
			for ( int j = q[i].l ; j <= q[i].r ; j ++ ) temp = max ( temp , tmp[a[j]] - j );
			ans[q[i].id] = temp;
			for ( int j = q[i].l ; j <= q[i].r ; j ++ ) tmp[a[j]] = 0;
			continue;
		}
		if ( lstblock ^ bel[q[i].l] )
		{
			for ( int j = l ; j <= r ; j ++ ) st[a[j]] = ed[a[j]] = 0;
			l = rr[bel[q[i].l]] + 1 , r = rr[bel[q[i].l]];
			res = 0; lstblock = bel[q[i].l];
		}
		while ( r < q[i].r ) { ++ r; if ( !st[a[r]] ) st[a[r]] = r; ed[a[r]] = r , res = max ( res , ed[a[r]] - st[a[r]] ); }
		int temp = res , _l = l;
		while ( q[i].l < _l ) { -- _l; if ( !temped[a[_l]] ) temped[a[_l]] = _l; tempst[a[_l]] = _l , res = max ( res , max ( temped[a[_l]] , ed[a[_l]] ) - tempst[a[_l]] ); }
		ans[q[i].id] = res;
		while ( _l < l ) tempst[a[_l]] = temped[a[_l]] = 0 , ++_l;
		res = temp;
	}
	for ( int i = 1 ; i <= m ; i ++ ) cout << ans[i] << endl;
	return 0;
}

P4147 玉蟾宫

先预处理每一个点能向上延伸多远的距离

那么我们对于每一行维护一个单调递增的单调栈栈中记录\(hei\)和\(wei\)两个数表示这个块的宽度和高度

每次将高度大于等于当前高度的栈顶都弹出每一次弹栈都统计答案记录 \(temp\) 表示能扩展的最大宽度即可

#include <bits/stdc++.h>
using namespace std;
#define mid (l+r>>1)
#define endl '\n'
#define inl inline
#define eb emplace_back
const int N = 1e3 + 5;
// char buf[1<<24] , *p1 , *p2;
// #define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<24,stdin),p1==p2)?EOF:*p1++)
#define getchar() cin.get();
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f ;
}

int n , m , pos[N][N] , ans;

struct que { int wei , hei; } sta[10000+5];

void solve ( int i )
{
	int top = 0 , temp = 0;
	sta[++top] = { 1 , pos[i][1] };
	for ( int j = 2 ; j <= m ; j ++ )
	{
		temp = 0;
		while ( sta[top].hei >= pos[i][j] && top )
		{
			temp += sta[top].wei;
			ans = max ( ans , sta[top--].hei * temp );
		}
		sta[++top] = { temp + 1 , pos[i][j] };
	}
	temp = 0;
	while ( top > 0 )
	{
		temp += sta[top].wei;
		ans = max ( ans , sta[top--].hei * temp );
	}
}

char ch;
signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	n = read() , m = read();
	for ( int i = 1 ; i <= n ; i ++ )
		for ( int j = 1 ; j <= m ; j ++ )
			cin >> ch , pos[i][j] = ( ch == 'F' ) ? pos[i-1][j] + 1 : 0;
	for ( int i = 1 ; i <= n ; i ++ ) solve ( i );
	cout << ans * 3 << endl;
	return 0;
}

posted @ 2023-08-14 10:43 Echo_Long 阅读(10) 评论(0) 编辑收藏举报

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Echo-Long

8.13 杂题

P1471 方差

P5906 【模板】回滚莫队&不删除莫队

P4147 玉蟾宫

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