Spoj 2916 can you answer these queries v line segment tree to find the maximum child segment and

Question link: Click the open link

Question:

T Test Data

N numbers

Q inquiries

For each query: [X1, Y1] [X2, y2]

Q:

int ans = -inf;for(int i = x1; i <= y1; i++)for(int j = max(x2, i); j <= y2; j++)ans = max(ans, query(i, j));

Ideas:

Query_l (int l, int R) calculates the maximum child segment and the left endpoint of L within the range [L, R.

Similar to gss3.

Classification to discuss the given interval [X1, Y1] [X2, y2]

1. If two intervals are separated (Y1 <X2), the (Y1, X2) part is required, and the other part is expanded as much as possible.

2. If Y1 = x2, this point in the middle is required. The other two sides are either deselected or expanded as much as possible.

3. When the interval is retained, we can discuss the interval and divide it into [X1, Y1), [Y1, x2], (X2, y2], then [I, j: There are four situations that fall into these three intervals.

The problem with gss3.

The other two cases are similar to 2.

# Include <cstdio> # include <iostream> # include <algorithm> # include <string. h> # include <math. h >#include <vector >#include <map> using namespace STD; # define n 10005 # define lson (x) tree [X]. L # define rson (x) tree [X]. R # define L (x) (x <1) # define R (x) (x <1 | 1) # define sum (x) tree [X]. sum # define max (x) tree [X]. max # define Lmax (x) tree [X]. lmax # define rmax (x) tree [X]. rmaxstruct node {int L, R; int mid () {return (L + r)> 1;} int Lmax, rmax, Max, sum ;} tree [n <2]; int n, a [n], Q; int query_sum (int l, int R, int ID) {If (L = lson (ID) & rson (ID) = r) return sum (ID); int mid = tree [ID]. mid (); If (mid <L) return query_sum (L, R, R (ID); else if (r <= mid) return query_sum (L, R, L (ID); else return query_sum (L, mid, L (ID) + query_sum (Mid + 1, R, R (ID ));} void push_up (int id) {Lmax (ID) = max (Lmax (L (ID), sum (L (ID) + Lmax (R (ID ))); rmax (ID) = max (rmax (R (ID), sum (R (ID) + rmax (L (ID); sum (ID) = sum (L (ID) + sum (R (ID); max (ID) = max (L (ID )), max (R (ID), rmax (L (ID) + Lmax (R (ID);} void updata_point (INT Val, int ID) {Lmax (ID) = rmax (ID) = max (ID) = sum (ID) = val;} void build (int l, int R, int ID) {lson (ID) = L; rson (ID) = r; If (L = r) {updata_point (A [L], ID); return ;} int mid = tree [ID]. mid (); Build (L, mid, L (ID); Build (Mid + 1, R, R (ID); push_up (ID );} int query_l (int l, int R, int ID) {If (L = lson (ID) & rson (ID) = r) return Lmax (ID ); int mid = tree [ID]. mid (); If (mid <L) return query_l (L, R, R (ID); else if (r <= mid) return query_l (L, R, L (ID); int LANs = query_l (L, mid, L (ID), Rans = query_l (Mid + 1, R, R (ID )); return max (LANs, sum (L (ID) + RANS);} int query_r (int l, int R, int ID) {If (L = lson (ID) & rson (ID) = r) return rmax (ID); int mid = tree [ID]. mid (); If (mid <L) return query_r (L, R, R (ID); else if (r <= mid) return query_r (L, R, L (ID); int LANs = query_r (L, mid, L (ID), Rans = query_r (Mid + 1, R, R (ID )); return max (RANS, sum (R (ID) + LANs);} int query_l (int l, int R, int ID) {If (L = lson (ID) & rson (ID) = r) return Lmax (ID); int mid = tree [ID]. mid (); If (mid <L) return query_l (L, R, R (ID); else if (r <= mid) return query_l (L, R, L (ID); int LANs = query_l (L, mid, L (ID), Rans = query_l (Mid + 1, R, R (ID )); return max (LANs, query_sum (L, mid, ID) + RANS);} int query_r (int l, int R, int ID) {If (L = lson (ID) & rson (ID) = r) return rmax (ID); int mid = tree [ID]. mid (); If (mid <L) return query_r (L, R, R (ID); else if (r <= mid) return query_r (L, R, L (ID); int LANs = query_r (L, mid, L (ID), Rans = query_r (Mid + 1, R, R (ID )); return max (RANS, query_sum (Mid + 1, R, ID) + LANs);} int query (int l, int R, int ID) {If (L = lson (ID) & rson (ID) = r) return max (ID); int mid = tree [ID]. mid (); If (mid <L) return query (L, R, R (ID); else if (r <= mid) return query (L, R, L (ID); int LANs = query (L, mid, L (ID), Rans = query (Mid + 1, R, R (ID )); int ans = max (LANs, RANS); Return max (ANS, query_r (L, mid, L (ID) + query_l (Mid + 1, R, R (ID);} void solve () {scanf ("% d", & N); For (INT I = 1; I <= N; I ++) scanf ("% d", & A [I]); Build (1, n, 1); scanf ("% d", & Q); While (Q --) {int x1, x2, Y1, Y2, ans; scanf ("% d", & X1, & Y1, & X2, & Y2 ); if (Y1 <X2) {ans = query_r (x1, Y1, 1) + query_l (X2, Y2, 1); If (Y1 + 1 <= x2-1) ans + = query_sum (Y1 + 1, x2-1, 1);} else if (Y1 = x2) {ans = query_sum (Y1, X2, 1 ); if (X1 <= y1-1) ans + = max (0, query_r (x1, y1-1, 1); If (X2 + 1 <= Y2) ans + = max (0, query_l (X2 + 1, Y2, 1);} else {ans = query (X2, Y1, 1); // I, J is in the middle of int TMP = query_r (x1, x2, 1) + query_l (Y1, Y2, 1); // I, j All in 2 side if (X2 + 1 <= y1-1) TMP + = query_sum (X2 + 1, y1-1, 1); ans = max (ANS, TMP ); TMP = query_l (X2, Y2, 1); If (X1 <= x2-1) TMP + = query_r (x1, x2-1, 1); ans = max (ANS, TMP ); TMP = query_r (x1, Y1, 1); If (Y1 + 1 <= Y2) TMP + = query_l (Y1 + 1, Y2, 1); ans = max (ANS, TMP);} printf ("% d \ n", ANS) ;}int main () {int t; scanf ("% d", & T ); while (t --) solve (); Return 0 ;}

Spoj 2916 can you answer these queries v line segment tree to find the maximum child segment and