Solution -「ZJOI 2016」「洛谷 P3352」线段树

(mathcal{Descrtiption})

  给定 ({a_n})，现进行 (m) 次操作，每次操作随机一个区间 ([l,r])，令其中元素全部变为区间最大值。对于每个 (i)，求所有可能操作方案最终得到的 (a_i) 之和。答案模 ((10^9+7))。

  (n,qle400)。

(mathcal{Solution})

  那什么我懒得写题解了就把草稿贴上来好了。（

[f(i,l,r,x):= ext{the operating ways that after }i ext{-th operation,}\ forall iin[l,r],a_ile x ext{ and }a_{l-1},a_{r+1}>x.\ f(i,l,r,x)=left(inom{l}{2}+inom{r-l+2}{2}+inom{n-r+1}{2} ight)f(i-1,l,r,x)\ +sum_{p<l}(p-1)f(i-1,p,r,x)+sum_{r<p}(n-p)f(i-1,l,p,x).\ ext{Thus, the answer for }i ext{ can be represented as }r_i ext{, where}\ egin{aligned} r_i&=sum_{x} xsum_{[l,r] i i}f(q,l,r,x)-f(q,l,r,x-1)\ &=sum_{[l,r] i i}sum_{x}-f(q,l,r,x)\ &=sum_{[l,r] i i}g(q,l,r)~~~~(g(i,l,r):=sum_{x}-f(i,l,r,x)). end{aligned}\ ext{Obviously, }g ext{'s expression is similar to }f ext{'s.}\ ext{Maintaining partial sum, this problem can be solved in }mathcal O(qn^2). ]

(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 )

const int MAXN = 400, MOD = 1e9 + 7, IINF = 0x3f3f3f3f;
int n, m, a[MAXN + 5];
int g[2][MAXN + 5][MAXN + 5], sum[MAXN + 5];

inline int tot( const int u ) { return ( u * ( u + 1ll ) >> 1 ) % MOD; }
inline int mul( const int u, const int v ) { return 1ll * u * v % MOD; }
inline int sub( int u, const int v ) { return ( u -= v ) < 0 ? u + MOD : u; }
inline int add( int u, const int v ) { return ( u += v ) < MOD ? u : u - MOD; }

int main() {
	scanf( "%d %d", &n, &m );
	rep ( i, 1, n ) scanf( "%d", &a[i] );
	a[0] = a[n + 1] = IINF;
	
	rep ( l, 1, n ) {
		int mxv = a[l];
		rep ( r, l, n ) {
			mxv = mxv < a[r] ? a[r] : mxv;
			int mnv = a[l - 1] < a[r + 1] ? a[l - 1] : a[r + 1];
			if ( mnv > mxv ) g[0][l][r] = sub( mxv, mnv < IINF ? mnv : 0 );
		}
	}
	
	for ( int sta = 1, i = 1; i <= m; sta ^= 1, ++i ) {
		memset( sum, 0, sizeof sum );
		rep ( l, 1, n ) rep ( r, l, n ) {
			g[sta][l][r] = add( sum[r], mul( add( add( tot( l - 1 ),
			  tot( r - l + 1 ) ), tot( n - r ) ), g[!sta][l][r] ) );
			sum[r] = add( sum[r], mul( l - 1, g[!sta][l][r] ) );
		}
		memset( sum, 0, sizeof sum );
		per ( r, n, 1 ) per ( l, r, 1 ) {
			g[sta][l][r] = add( g[sta][l][r], sum[l] );
			sum[l] = add( sum[l], mul( n - r, g[!sta][l][r] ) );
		}
	}
	
	rep ( i, 1, n ) {
		int ans = 0;
		rep ( l, 1, i ) rep ( r, i, n ) ans = add( ans, g[m & 1][l][r] );
		printf( "%d%c", ans, i < n ? ' ' : '
' );
	}
	return 0;
}