Poj 3347 kadj squares (ry + line segment tree)

Question:

Place squares as shown in the figure. Output the squares from the top to the bottom.

Question: 1: because all the line segments have a 45/135-degree angle, all data is larger than SQRT (2) and can be processed as integers. 2: When we use a line segment tree to handle the entire point, we need to pay attention to the following situations because we can only process the entire point.

That is, 3rd can only be seen from the leftmost point. However, 2nd marks this point as 2, and the result is 3, which is invisible. To avoid this problem, we can multiply the left and right points of each square by 2, and then subtract 1 from the right vertex (similar to the processing of the open and closed intervals)

# Include <cmath> # include <cstdio> # include <cstdlib> # include <algorithm> # include <cstring> using namespace STD; # define L (u) <1) # define R (u) <1 | 1) struct square {int B, l, R, Len, ID, flag ;} s [500]; struct node {int L, R, flag;} node [10000]; bool mark [500]; void build (int u, int L, int R) {node [u]. L = L; node [u]. R = r; node [u]. flag = 0; If (L = r) return; int mid = (L + r)> 1; build (L (u ), L, mid); Build (R (u), Mid + 1, R);} void Update (int u, int L, int R, int F) {If (node [u]. L = L & node [u]. R = r) {node [u]. flag = f; return;} If (node [u]. flag! = 0) {node [L (u)]. flag = node [u]. flag; node [R (u)]. flag = node [u]. flag; node [u]. flag = 0;} int mid = (node [u]. L + node [u]. r)> 1; if (r <= mid) Update (L (u), L, r, f); else if (L> mid) update (R (u), L, r, f); else {Update (L (u), L, mid, f); Update (R (u ), mid + 1, r, f) ;}} void query (int u) {If (node [u]. flag! = 0) {mark [node [u]. flag] = true; return;} If (node [u]. L = node [u]. r) return; query (L (u); query (R (u);} bool CMP (const square & A, const square & B) {return. len <= B. len;} int main () {int N; while (scanf ("% d", & N) {int I, j, TMP, T = 0; for (I = 1; I <= N; I ++) {scanf ("% d", & S [I]. len); s [I]. id = I; TMP = 0; if (I = 1) s [I]. B = s [I]. len; else {for (j = I-1; j> = 1; j --)// Enumerate all the rectangles before I to compact the rectangle I (s [I]. len> S [J]. len) TMP = max (TMP, s [J]. B + s [J]. len * 2); else {TMP = max (TMP, s [J]. B + s [I]. len * 2); break;} If (j <1 & S [I]. len> = TMP) s [I]. B = s [I]. len; else s [I]. B = TMP;} s [I]. L = (s [I]. b-s [I]. len) * 2; // note !!! S [I]. r = (s [I]. B + s [I]. Len) * 2-1 ;//!!! T = max (s [I]. r, T);} Sort (S + 1, S + 1 + N, CMP); Build (1, 0, T); memset (mark, 0, sizeof (Mark); for (I = 1; I <= N; I ++) Update (1, s [I]. l, s [I]. r, s [I]. ID); query (1); for (I = 1; I <= N; I ++) if (MARK [I]) printf ("% d", I ); printf ("\ n");} return 0 ;}