Zoj 2301/HDU 1199 color the ball discretization + continuous maximum sum of Line Segment trees

Question: Let's arrange n balls in a row. They are black at first. Now, let's perform some operations (L, R, color) and color the results in the [L, R] interval, 'W' is white, and 'B' is black. Ask the start point and end point of the longest white interval.

Solution: First discretization, in order to prevent discrete error, not only will l, r discrete, but also add L + 1, L-1, R + 1, R-1 discrete together, in this way, no problem will occur. Create a line segment tree and maintain four values:

1. Col range color 0 indicates black 1 indicates white-1 indicates no mark

2. The length of the maximum white interval in the maxi interval. Because the white is expressed as 1, the maximum white Interval Length is the maximum continuous and

3. Maximum length of the Lmax white interval starting from the left end of the Interval

4. Maximum white Interval Length of rmax starting from the right end of the Interval

Then update and query, which is the maximum continuous and no difference from the common interval.

Code:

# Include <iostream> # include <cmath> # include <iostream> # include <cstdio> # include <cstring> # include <cstdlib> # include <cmath> # include <algorithm> # include <map> using namespace STD; # define n 14007 struct node {int Lmax, rmax, Maxi, Col;} tree [4 * n]; int num [N], X [N]; int L [2005], R [2005]; char ss [2005] [3]; Map <int, int> MP; void pushup (int l, int R, int RT) {tree [RT]. lmax = tree [2 * RT]. lmax; tree [RT]. rmax = tree [2 * RT + 1 ]. Rmax; tree [RT]. maxi = max (tree [2 * RT]. maxi, tree [2 * RT + 1]. maxi), tree [2 * RT]. rmax + tree [2 * RT + 1]. lmax); int mid = (L + r)/2; int L = x [Mid]-x L-1]; // true length int r = x [R]-X [Mid]; If (tree [2 * RT]. lmax = L) tree [RT]. lmax + = tree [2 * RT + 1]. lmax; If (tree [2 * RT + 1]. rmax = r) tree [RT]. rmax + = tree [2 * RT]. rmax;} void Pushdown (int l, int R, int RT) {If (tree [RT]. col! =-1) {tree [2 * RT]. col = tree [2 * RT + 1]. col = tree [RT]. col; int mid = (L + r)/2; int L = x [Mid]-x L-1]; int r = x [R]-X [Mid]; tree [2 * RT]. maxi = tree [2 * RT]. lmax = tree [2 * RT]. rmax = L * tree [RT]. col; tree [2 * RT + 1]. maxi = tree [2 * RT + 1]. lmax = tree [2 * RT + 1]. rmax = r * tree [RT]. col; tree [RT]. col =-1 ;}} void build (int l, int R, int RT) {tree [RT]. maxi = tree [RT]. lmax = tree [RT]. rmax = 0; tree [RT]. col =-1; if (L = r) Return; int mid = (L + r)/2; build (L, mid, 2 * RT); Build (Mid + 1, R, 2 * RT + 1 );} void Update (int l, int R, int AA, int BB, int Col, int RT) {If (AA <= L & BB> = r) {tree [RT]. col = Col; tree [RT]. maxi = tree [RT]. lmax = tree [RT]. rmax = Col * (X [R]-L-1]); return;} Pushdown (L, R, RT); int mid = (L + r)/2; if (AA <= mid) Update (L, mid, AA, BB, Col, 2 * RT); If (BB> mid) Update (Mid + 1, R, AA, BB, Col, 2 * RT + 1); pushup (L, R, RT);} int query (Int L, int R, int RT) {If (L = r) return X [l]; int mid = (L + r)/2; Pushdown (L, R, RT); If (tree [2 * RT]. maxi = tree [1]. maxi) // tree [1] Not tree [RT] Return query (L, mid, 2 * RT); else if (tree [2 * RT]. rmax + tree [2 * RT + 1]. lmax = tree [1]. maxi) return X [Mid]-tree [2 * RT]. rmax + 1; else return query (Mid + 1, R, 2 * RT + 1);} int main () {int N, I, J, K; while (scanf ("% d", & N )! = EOF) {MP. clear (); for (I = 1; I <= N; I ++) {scanf ("% d % s", & L [I], & R [I], ss [I]); num [6 * I-5] = L [I]-1; num [6 * I-4] = L [I]; num [6 * I-3] = L [I] + 1; num [6 * I-2] = R [I]-1; num [6 * I-1] = R [I]; num [6 * I] = R [I] + 1;} Sort (Num + 1, num + 6 * n + 1); int ind = unique (Num + 1, num + 6 * n + 1)-num-1; int now = 0; X [0] = 0; for (I = 1; I <= ind; I ++) {x [++ now] = num [I]; MP [num [I] = now;} build (1, now, 1); for (I = 1; I <= N; I ++) {int Ka = MP [L [I]; int kb = MP [R [I]; if (ss [I] [0] = 'W') Update (1, now, ka, kb, 1, 1); else update (1, now, ka, kb, 0, 1);} If (tree [1]. maxi <= 0) puts ("Oh, my God"); else {int left = query (1, now, 1); int right = left + tree [1]. maxi-1; printf ("% d \ n", left, right) ;}} return 0 ;}

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Zoj 2301/HDU 1199 color the ball discretization + continuous maximum sum of Line Segment trees