Poj 2528 mayor's posters [discretization + line segment tree]

Title: poj 2528 mayor's posters

Question: Add a poster with a very long length of AI to the wall. Because some of them will overwrite it, find the number of posters that can be seen at last.

Analysis: The question is exactly the same as that of poj2777, and the method is the same, but this should be discretization, And the array should be larger. At least 2 times.

The C ++ pair class is used for discretization, which is relatively easy to use.

Code:

# Include <iostream> # include <algorithm> # include <utility> # include <cstring> # include <cstdio> using namespace STD; const int n = 25000; pair <int, int> P [2 * n]; pair <int, int> V [N]; int N; int comp (pair <int, int> A, pair <int, int> B) {if (. first! = B. first) return. first <B. first;} int CMP (pair <int, int> A, pair <int, int> B) {if (. second! = B. Second) return a. Second <B. Second; If (A. First! = B. first) return. first <B. first;} int discrete () // discretization {for (INT I = 0; I <n; I ++) {int X, Y; scanf ("% d", & X, & Y); P [I * 2] = make_pair (X, I ); P [I * 2 + 1] = make_pair (Y, I);} Sort (p, p + 2 * n, comp); int CNT = 2, TMP = P [0]. first; P [0]. first = 1; for (INT I = 1; I <n * 2; I ++) {If (P [I]. first = TMP) {TMP = P [I]. first; P [I]. first = (cnt-1);} else {TMP = P [I]. first; P [I]. first = CNT ++ ;}} sort (p, p + 2 * n, CMP); For (INT I = 0; I <2 * n; I + = 2) V [I /2] = make_pair (P [I]. first, P [I + 1]. first); CNT --; return CNT;} struct node {int L, R; long num;}; node tree [4 * n]; int vis [N * 2]; void build (int l, int R, int o) {tree [O]. L = L; tree [O]. R = r; tree [O]. num = 1; if (L = r) return; int mid = (L + r)/2; build (L, mid, O * 2 ); build (Mid + 1, R, O * 2 + 1);} void Update (int l, int R, int T, int o) {If (tree [O]. L = L & tree [O]. R = r) {tree [O]. num = T; return;} If (tree [O]. num = T) return; If (Tree [O]. num! =-1) {tree [2 * o]. num = tree [O]. num; tree [2 * O + 1]. num = tree [O]. num; tree [O]. num =-1;} int mid = (tree [O]. L + tree [O]. r)> 1; if (r <= mid) Update (L, R, T, O + O); else if (L> mid) Update (L, R, t, O + 1); else {Update (L, mid, T, O * 2); Update (Mid + 1, R, T, O * 2 + 1) ;}} void query (int l, int R, int o) {If (tree [O]. num! =-1) {vis [tree [O]. num] = 1; return;} int mid = (tree [O]. L + tree [O]. r)> 1; if (r <= mid) query (L, R, O + O); else if (L> mid) query (L, R, O + 1); else {query (L, mid, O * 2); query (Mid + 1, R, O * 2 + 1 );}} int main () {int t; scanf ("% d", & T); While (t --) {scanf ("% d", & N ); int CNT = discrete (); Build (1, CNT, 1); For (INT I = 0; I <n; I ++) {Update (V [I]. first, V [I]. second, I, 1);} memset (VIS, 0, sizeof (VIS); query (1, CNT, 1); int ans = 0; for (INT I = 0; I <= CNT; I ++) if (vis [I]) {// printf ("-- % d", I ); ans ++;} printf ("% d \ n", ANS); memset (p, 0, sizeof (p); memset (v, 0, sizeof (v);} return 0 ;}