SP1716 GSS3 - Can you answer these queries III

题目链接

问题分析

解法一

简单的一道线段树维护\(lmax,rmax,max,sum\)即可。

解法二

考虑动态DP。

\[\begin{aligned} \left [ \begin{matrix} a_i&-\infty&a_i\\ a_i&0&a_i\\-\infty & -\infty &0 \end{matrix}\right ] \left [\begin{matrix} f_{i-1}\\ g_{i-1}\\ 0 \end{matrix}\right] = \left [\begin{aligned} f_i\\ g_i\\ 0 \end{aligned}\right] \end{aligned} \]

然后线段树维护即可。

参考程序

解法一

#include <bits/stdc++.h>
using namespace std;

const int Maxn = 50010;
struct node {
	int LeftMax, RightMax, Max, Sum;
	node() {}
	node( int _LeftMax, int _RightMax, int _Max, int _Sum) : 
		LeftMax( _LeftMax ), RightMax( _RightMax ), Max( _Max ), Sum( _Sum ) {}
	inline node operator + ( const node Other ) const {
		return node( max( LeftMax, Sum + Other.LeftMax ), 
				max( Other.RightMax, Other.Sum + RightMax ),
				max( max( Max, Other.Max ), RightMax + Other.LeftMax ),
				Sum + Other.Sum );
	}
	inline void Give( int x ) {
		LeftMax = RightMax = Max = Sum = x;
		return;
	}
};
node Tree[ Maxn << 2 ];
int n, m, A[ Maxn ];

void Build( int Index, int Left, int Right ) {
	if( Left == Right ) {
		Tree[ Index ].Give( A[ Left ] );
		return;
	}
	int Mid = ( Left + Right ) >> 1;
	Build( Index << 1, Left, Mid );
	Build( Index << 1 | 1, Mid + 1, Right );
	Tree[ Index ] = Tree[ Index << 1 ] + Tree[ Index << 1 | 1 ];
	return;
}

void Change( int Index, int Left, int Right, int Pos, int Key ) {
	if( Left == Right ) {
		Tree[ Index ].Give( Key );
		return;
	}
	int Mid = ( Left + Right ) >> 1;
	if( Pos <= Mid ) Change( Index << 1, Left, Mid, Pos, Key );
	if( Pos > Mid ) Change( Index << 1 | 1, Mid + 1, Right, Pos, Key );
	Tree[ Index ] = Tree[ Index << 1 ] + Tree[ Index << 1 | 1 ];
	return;
}

node Query( int Index, int Left, int Right, int L, int R ) {
	if( L <= Left && Right <= R ) return Tree[ Index ];
	int Mid = ( Left + Right ) >> 1;
	if( R <= Mid ) return Query( Index << 1, Left, Mid, L, R );
	if( L > Mid ) return Query( Index << 1 | 1, Mid + 1, Right, L, R );
	return Query( Index << 1, Left, Mid, L, R ) + Query( Index << 1 | 1, Mid + 1, Right, L, R );
}

int main() {
	scanf( "%d", &n );
	for( int i = 1; i <= n; ++i ) scanf( "%d", A + i );
	Build( 1, 1, n );
	scanf( "%d", &m );
	for( int i = 1; i <= m; ++i ) {
		int Opt, l, r;
		scanf( "%d%d%d", &Opt, &l, &r );
		if( Opt == 0 ) Change( 1, 1, n, l, r );
		if( Opt == 1 ) printf( "%d\n", Query( 1, 1, n, l, r ).Max );
	}
	return 0;
}

解法二

#include <bits/stdc++.h>
using namespace std;

const int Maxn = 50010;
const int INF = 1000000000;
struct matrix {
	int A[ 3 ][ 3 ];
	matrix() {
		for( int i = 0; i < 3; ++i )
			for( int j = 0; j < 3; ++j )
				A[ i ][ j ] = -INF;
		return;
	}
	matrix( int x ) {
		A[ 0 ][ 0 ] = A[ 0 ][ 2 ] = A[ 1 ][ 0 ] = A[ 1 ][ 2 ] = x;
		A[ 0 ][ 1 ] = A[ 2 ][ 0 ] = A[ 2 ][ 1 ] = -INF;
		A[ 1 ][ 1 ] = A[ 2 ][ 2 ] = 0;
		return;
	}
	inline matrix operator * ( const matrix Other ) const {
		matrix Ans = matrix();
		for( int i = 0; i < 3; ++i )
			for( int j = 0; j < 3; ++j )
				for( int k = 0; k < 3; ++k )
					Ans.A[ i ][ j ] = max( Ans.A[ i ][ j ], A[ i ][ k ] + Other.A[ k ][ j ] );
		return Ans;
	}
};
int n, m, A[ Maxn ];
matrix Tree[ Maxn << 2 ];

void Build( int Index, int Left, int Right ) {
	if( Left == Right ) {
		Tree[ Index ] = matrix( A[ Left ] );
		return;
	}
	int Mid = ( Left + Right ) >> 1;
	Build( Index << 1, Left, Mid );
	Build( Index << 1 | 1, Mid + 1, Right );
	Tree[ Index ] = Tree[ Index << 1 ] * Tree[ Index << 1 | 1 ];
	return;
}

void Change( int Index, int Left, int Right, int Pos, int Key ) {
	if( Left == Right ) {
		Tree[ Index ] = matrix( Key );
		return;
	}
	int Mid = ( Left + Right ) >> 1;
	if( Pos <= Mid ) Change( Index << 1, Left, Mid, Pos, Key );
	if( Pos > Mid ) Change( Index << 1 | 1, Mid + 1, Right, Pos, Key );
	Tree[ Index ] = Tree[ Index << 1 ] * Tree[ Index << 1 | 1 ];
	return;
}

matrix Query( int Index, int Left, int Right, int L, int R ) {
	if( L <= Left && Right <= R ) return Tree[ Index ];
	int Mid = ( Left + Right ) >> 1;
	if( R <= Mid ) return Query( Index << 1, Left, Mid, L, R );
	if( L > Mid ) return Query( Index << 1 | 1, Mid + 1, Right, L, R );
	return Query( Index << 1, Left, Mid, L, R ) * Query( Index << 1 | 1, Mid + 1, Right, L, R );
}

int main() {
	scanf( "%d", &n );
	for( int i = 1; i <= n; ++i ) scanf( "%d", &A[ i ] );
	Build( 1, 1, n );
	scanf( "%d", &m );
	for( int i = 1; i <= m; ++i ) {
		int Opt, x, y;
		scanf( "%d%d%d", &Opt, &x, &y );
		if( Opt == 0 ) Change( 1, 1, n, x, y );
		if( Opt == 1 ) printf( "%d\n", Query( 1, 1, n, x, y ).A[ 1 ][ 2 ] );
	}
	return 0;
}
posted @ 2019-09-10 20:19  chy_2003  阅读(...)  评论(...编辑  收藏