Solution -「多校联训」古老的序列问题
\(\mathcal{Description}\)
Link.
给定序列 \(\{a_n\}\)，和 \(q\) 次形如 \([L,R]\) 的询问，每次回答
\[\sum_{[l,r]\subseteq [L,R]}\min_{i=l}^r\{a_i\}\cdot\max_{i=l}^r\{a_i\}\pmod{10^9+7}.
\]
\(n,q\le10^5\)。
\(\mathcal{Solution}\)
瞬间联想到 这道题，尝试把询问挂到猫树上分治处理。对于分治区间 \([l,r]\)，令其中点为 \(p\)，考虑离线处理挂在它身上的询问。
而此时，我们需要用扫描线进行进一步转化：左端点 \(i\) 从 \(p\) 向 \(l\) 扫描，维护 \(j=p+1..r\) 的答案。考虑 \(i\) 确定时，\(j\in(p,r]\) 形成的 \([i,j]\) 的权值有以下四种：
- 最小值、最大值在 \([i,p]\)；
- 最小值在 \([i,p]\)，最大值在 \((p,r]\)；
- 最大值在 \([i,p]\)，最小值在 \((p,r]\)；
- 最小值、最大值在 \((p,r]\)。
简直和上面那题一模一样呢。发现贡献无非是左边的最小/最大值等形式的系数乘上右边相同形式的系数，询问时即求右边贡献的前缀和。可以用四科线段树维护四种类型的贡献。复杂度 \(\mathcal O((q+n\log n)\log n)\)。
Ummm... 有 \(\mathcal O(n\log n)\) 的做法，大概是只用扫描线，然后矩阵线段树维护历史和，见 祂的博客。
\(\mathcal{Code}\)
/*~Rainybunny~*/
#include <bits/stdc++.h>
#define rep( i, l, r ) for ( int i = l, rep##i = r; i <= rep##i; ++i )
#define per( i, r, l ) for ( int i = r, per##i = l; i >= per##i; --i )
inline char fgc() {
static char buf[1 << 17], *p = buf, *q = buf;
return p == q && ( q = buf + fread( p = buf, 1, 1 << 17, stdin ), p == q )
? EOF : *p++;
}
inline int rint() {
int x = 0, s = fgc();
for ( ; s < '0' || '9' < s; s = fgc() );
for ( ; '0' <= s && s <= '9'; s = fgc() ) x = x * 10 + ( s ^ '0' );
return x;
}
inline void wint( const int x ) {
if ( 9 < x ) wint( x / 10 );
putchar( x % 10 ^ '0' );
}
const int MAXN = 1e5, MOD = 1e9 + 7;
int n, q, a[MAXN + 5], ans[MAXN + 5];
inline int imin( const int u, const int v ) { return u < v ? u : v; }
inline int imax( const int u, const int v ) { return u < v ? v : u; }
inline int mul( const int u, const int v ) { return 1ll * u * v % MOD; }
inline int add( int u, const int v ) { return ( u += v ) < MOD ? u : u - MOD; }
inline void addeq( int& u, const int v ) { ( u += v ) >= MOD && ( u -= MOD ); }
struct Atom {
int l, r, id;
inline bool operator < ( const Atom& t ) const { return l < t.l; }
};
std::vector<Atom> qbuc[MAXN << 2];
int all[MAXN << 2];
inline void hang( const int u, const int l, const int r,
const int ql, const int qr, const int qid ) {
if ( ql <= l && r <= qr ) return qbuc[u].push_back( { l, r, qid } );
int mid = l + r >> 1;
if ( qr <= mid ) hang( u << 1, l, mid, ql, qr, qid );
else if ( mid < ql ) hang( u << 1 | 1, mid + 1, r, ql, qr, qid );
else {
qbuc[u].push_back( { ql, qr, qid } );
hang( u << 1, l, mid, ql, mid, qid );
hang( u << 1 | 1, mid + 1, r, mid + 1, qr, qid );
}
}
struct SegmentTree { // how to replace it with BIT?
int sum[MAXN << 2], coe[MAXN << 2], tag[MAXN << 2];
inline void pushad( const int u, const int v ) {
addeq( tag[u], v ), addeq( sum[u], mul( coe[u], v ) );
}
inline void pushdn( const int u ) {
if ( tag[u] ) {
pushad( u << 1, tag[u] ), pushad( u << 1 | 1, tag[u] );
tag[u] = 0;
}
}
inline void pushup( const int u ) {
sum[u] = add( sum[u << 1], sum[u << 1 | 1] );
}
inline void build( const int u, const int l, const int r, const int* c ) {
sum[u] = tag[u] = 0;
if ( l == r ) return void( coe[u] = c == NULL ? 1 : c[l] );
int mid = l + r >> 1;
build( u << 1, l, mid, c ), build( u << 1 | 1, mid + 1, r, c );
coe[u] = add( coe[u << 1], coe[u << 1 | 1] );
}
inline void modify( const int u, const int l, const int r,
const int ml, const int mr, const int v ) {
// if ( l > r ) return ; // well, I'll promise it.
if ( ml <= l && r <= mr ) return pushad( u, v );
int mid = l + r >> 1; pushdn( u );
if ( ml <= mid ) modify( u << 1, l, mid, ml, mr, v );
if ( mid < mr ) modify( u << 1 | 1, mid + 1, r, ml, mr, v );
pushup( u );
}
inline int query( const int u, const int l, const int r,
const int ql, const int qr ) {
if ( ql <= l && r <= qr ) return sum[u];
int mid = l + r >> 1, ret = 0; pushdn( u );
if ( ql <= mid ) addeq( ret, query( u << 1, l, mid, ql, qr ) );
if ( mid < qr ) addeq( ret, query( u << 1 | 1, mid + 1, r, ql, qr ) );
return ret;
}
} sgt[4];
inline void solve( const int u, const int l, const int r ) {
/* bound check and divide down. */
auto& qry( qbuc[u] );
if ( l == r ) {
all[u] = mul( a[l], a[l] );
for ( const auto& x: qry ) addeq( ans[x.id], all[u] );
return ;
}
int mid = l + r >> 1;
solve( u << 1, l, mid ), solve( u << 1 | 1, mid + 1, r );
all[u] = add( all[u << 1], all[u << 1 | 1] );
/* initialize some information. */
static int x[MAXN + 5], y[MAXN + 5], xy[MAXN + 5]; // x->min, y->max.
x[mid] = y[mid] = a[mid];
per ( i, mid - 1, l ) {
x[i] = imin( a[i], x[i + 1] ), y[i] = imax( a[i], y[i + 1] );
}
x[mid + 1] = y[mid + 1] = a[mid + 1];
rep ( i, mid + 2, r ) {
x[i] = imin( a[i], x[i - 1] ), y[i] = imax( a[i], y[i - 1] );
}
rep ( i, l, r ) xy[i] = mul( x[i], y[i] );
#define R 1, 1, r - mid
sgt[0].build( R, NULL );
sgt[1].build( R, x + mid );
sgt[2].build( R, y + mid );
sgt[3].build( R, xy + mid );
/* finally begin to solve queries. */
std::sort( qry.begin(), qry.end() );
for ( int i = mid, j = int( qry.size() ) - 1,
px = mid + 1, py = mid + 1; i >= l; --i ) {
while ( px <= r && x[i] <= x[px] ) ++px;
while ( py <= r && y[i] >= y[py] ) ++py;
int pl = imin( px, py ), pr = imax( px, py );
if ( mid + 1 < pl ) sgt[0].modify( R, 1, pl - mid - 1, xy[i] );
if ( px < py ) sgt[1].modify( R, px - mid, py - mid - 1, y[i] );
if ( py < px ) sgt[2].modify( R, py - mid, px - mid - 1, x[i] );
if ( pr <= r ) sgt[3].modify( R, pr - mid, r - mid, 1 );
while ( ~j && qry[j].l == i ) {
int qr = qry[j].r - mid;
addeq( ans[qry[j].id], add(
add( sgt[0].query( R, 1, qr ), sgt[1].query( R, 1, qr ) ),
add( sgt[2].query( R, 1, qr ), sgt[3].query( R, 1, qr ) ) ) );
if ( i == l && qry[j].r == r ) addeq( ans[qry[j].id], all[u] );
--j;
}
}
/* update all[u] with contribution in current section. */
addeq( all[u], add(
add( sgt[0].query( R, 1, r - mid ), sgt[1].query( R, 1, r - mid ) ),
add( sgt[2].query( R, 1, r - mid ), sgt[3].query( R, 1, r - mid ) ) ) );
#undef R
}
int main() {
freopen( "sequence.in", "r", stdin );
freopen( "sequence.out", "w", stdout );
n = rint(), q = rint();
rep ( i, 1, n ) a[i] = rint();
rep ( i, 1, q ) {
int l = rint(), r = rint();
hang( 1, 1, n, l, r, i );
}
solve( 1, 1, n );
rep ( i, 1, q ) wint( ans[i] ), putchar( '\n' );
return 0;
}
