【SPOJ】2916 Can you answer these queries V 线段树

传送门：【SPOJ】2916 Can you answer these queries V

题目分析：仔细一点就行了。。

共分两种情况：

1.y1 <= x2，此时[ y1 , x2 ]内的数一定要取，两边看情况。

2.x2 < y1，此时分成三个区间[ x1 , x2 ] , [ x2 , y1 ] , [ y1 , y2 ]，然后一共有三种选择：

1.[ x2 , y1 ]中选取最大连续和

2.[ x1 , x2 ]中选取最大右连续和，[ x2 , y2 ]中选取最大左连续和

3.[ x1 , y1 ]中选取最大右连续和，[ y1 , y2 ]中选取最大左连续和

代码如下：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

#define REP( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )
#define FOR( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define REV( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define CLR( a , x ) memset ( a , x , sizeof a )

#define ls ( o << 1 )
#define rs ( o << 1 | 1 )
#define lson ls , l , m
#define rson rs , m + 1 , r
#define mid ( ( l + r ) >> 1 )
#define root 1 , 1 , n
#define rt o , l , r

const int MAXN = 10005 ;

int maxv[MAXN << 2] , lmax[MAXN << 2] , rmax[MAXN << 2] , num[MAXN] , sum[MAXN] ;

void build ( int o , int l , int r ) {
	if ( l == r ) {
		maxv[o] = lmax[o] = rmax[o] = num[l] ;
		return ;
	}
	int m = mid ;
	build ( lson ) , build ( rson ) ;
	lmax[o] = max ( lmax[ls] , sum[m] - sum[l - 1] + lmax[rs] ) ;
	rmax[o] = max ( rmax[rs] , sum[r] - sum[m] + rmax[ls] ) ;
	maxv[o] = max ( max ( maxv[ls] , maxv[rs] ) , rmax[ls] + lmax[rs] ) ;
}

int Lquery ( int L , int R , int o , int l , int r ) {
	if ( L > R ) return 0 ;
	if ( L <= l && r <= R ) return lmax[o] ;
	int m = mid ;
	if ( R <= m ) return Lquery ( L , R , lson ) ;
	if ( m <  L ) return Lquery ( L , R , rson ) ;
	return max ( Lquery ( L , m , lson ) , Lquery ( m + 1 , R , rson ) + sum[m] - sum[L - 1] ) ;
}

int Rquery ( int L , int R , int o , int l , int r ) {
	if ( L > R ) return 0 ;
	if ( L <= l && r <= R ) return rmax[o] ;
	int m = mid ;
	if ( R <= m ) return Rquery ( L , R , lson ) ;
	if ( m <  L ) return Rquery ( L , R , rson ) ;
	return max ( Rquery ( m + 1 , R , rson ) , Rquery ( L , m , lson ) + sum[R] - sum[m] ) ;
}

int query ( int L , int R , int o , int l , int r ) {
	if ( L > R ) return 0 ;
	if ( L <= l && r <= R ) return maxv[o] ;
	int m = mid ;
	if ( R <= m ) return query ( L , R , lson ) ;
	if ( m <  L ) return query ( L , R , rson ) ;
	int ans = max ( query ( L , m , lson ) , query ( m + 1 , R , rson ) ) ;
	return max ( ans , Rquery ( L , m , lson ) + Lquery ( m + 1 , R , rson ) ) ;
}

void solve () {
	int n , m ;
	int x1 , x2 , y1 , y2 ;
	scanf ( "%d" , &n ) ;
	FOR ( i , 1 , n ) scanf ( "%d" , &num[i] ) ;
	FOR ( i , 1 , n ) sum[i] = num[i] + sum[i - 1] ;
	build ( root ) ;
	scanf ( "%d" , &m ) ;
	while ( m -- ) {
		scanf ( "%d%d%d%d" , &x1 , &y1 , &x2 , &y2 ) ;
		int ans = 0 ;
		if ( y1 <= x2 ) {
			ans = Rquery ( x1 , y1 , root ) + Lquery ( x2 , y2 , root ) + sum[x2 - 1] - sum[y1] ;
		} else {
			ans = query ( x2 , y1 , root ) ;//1
			ans = max ( ans , Rquery ( x1 , x2 , root ) + Lquery ( x2 , y2 , root ) - num[x2] ) ;//2
			ans = max ( ans , Rquery ( x1 , y1 , root ) + Lquery ( y1 , y2 , root ) - num[y1] ) ;//3
		}
		printf ( "%d\n" , ans ) ;
	}	
}

int main () {
	int T ;
	scanf ( "%d" , &T ) ;
	while ( T -- ) solve () ;
	return 0 ;
}