Problem solving on both sides of coverage

Line Segment tree + Spanning Tree dyeing, Standard O (nlogn), no longer have to worry about being stuck... [Cpp] # include <cstdio> # include <cmath> # include <cstdlib> # include <cstring>/* # include <ctime> # include <cctype> # include <map> # include <set> # include <string> # include <queue> # include <iostream> # include <fstream> */# include <algorithm> using namespace std; # ifdef WIN32 # define fmt64 "% I64d" # else # define fmt64 "% lld" # endif # define PI M_PI # define oo 0x13131313 # define PB push_back # define PO pop_back # define MP make_pair # define iter iterator # define fst first # define snd second # define cstr (). c_str () # define FOR (I, j, k) for (I = (j); I <= (k); ++ I) # define ROF (I, j, k) for (I = (j); I >= (k); -- I) # define FER (e, d, u) for (e = d [u]; e; e = e-> n) # define FRE (I, a) for (I = (). begin (); I! = (). End (); ++ I) typedef unsigned int uint; typedef long int64; typedef unsigned long uint64; typedef long double real; template <class T> inline bool minim (T & a, const T & B) {return B <? A = B, 1: 0;} template <class T> inline bool maxim (T & a, const T & B) {return B>? A = B, 1: 0 ;}template <class T> inline T sqr (const T & a) {return a * a ;}# define maxn 300005 # define updatex (p) (p-> x = ll [p-> s-> x] <ll [p-> t-> x]? P-> s-> x: p-> t-> x) # define updatey (p) (p-> y = rr [p-> s-> y]> rr [p-> t-> y]? P-> s-> y: p-> t-> y) int n, m; int ll [maxn], rr [maxn]; int lx [maxn], rx [maxn]; int ly [maxn], ry [maxn]; struct node {node * s, * t, * f; int l, r, x, y ;}; node ns [maxn * 2], * nt = ns, * root, * bk [maxn]; int pt, lt, rt; char c [maxn]; int p [maxn], q [maxn], hd, tl; node * build (int l, int r) {node * p = ++ nt; p-> l = l, p-> r = r; if (l = r) return bk [l] = p; (p-> s = build (l, (l + r)> 1)-> f = p; return (p-> t = Build (l + r)> 1) + 1, r)-> f = p;} void insertx (int u) {node * p = bk [rr [u]; lx [p-> x] = u, rx [u] = p-> x, p-> x = u; for (p = p-> f; p = p-> f) updatex (p);} void inserty (int u) {node * p = bk [ll [u]; ly [p-> y] = u, ry [u] = p-> y, p-> y = u; for (p = p-> f; p = p-> f) updatey (p);} void remove (int u) {node * p = bk [rr [u]; bool flag = u = p-> x; rx [lx [u] = rx [u], lx [rx [u] = lx [u]; if (flag) {p-> X = rx [u]; for (p = p-> f; p = p-> f) updatex (p );} p = bk [ll [u], flag = u = p-> y; ry [ly [u] = ry [u], ly [ry [u] = ly [u]; if (flag) {p-> y = ry [u]; for (p = p-> f; p; p = p-> f) updatey (p) ;}} void query (node * p) {if (lt <p-> l & p-> r <rt) {for (; ll [p-> x] <lt; remove (p-> x )) c [q [++ tl] = p-> x] =-c [pt]; for (; rr [p-> y]> rt; remove (p-> y) c [q [++ tl] = p-> y] =-c [pt];} else {if (lt <p-> s-> r & (Ll [p-> s-> x] <lt | rr [p-> s-> y]> rt) query (p-> s ); if (rt> p-> t-> l & (ll [p-> t-> x] <lt | rr [p-> t-> y]> rt )) query (p-> t) ;}} int a [maxn], B [maxn]; int tx, ty; int top, stk [maxn]; bool cmp (int I, int j) {return ll [I] <ll [j] | (ll [I] = ll [j] & rr [I]> rr [j]);} void solve (int * a, int n) {sort (a + 1, a + n + 1, cmp); stk [top = 1] = a [1]; for (int I = 2; I <= n; ++ I) {for (; top & rr [A [I]> rr [stk [top]; -- top) if (ll [a [I] <rr [stk [top]) puts ("NIE"), exit (0) ;}} void bfs (int S) {for (c [q [hd = tl = 1] = S] = 1; hd <= tl; ++ hd) remove (pt = q [hd]), lt = ll [pt], rt = rr [pt], query (root ); tx = ty = 0; for (int I = 1; I <= tl; ++ I) if (~ C [q [I]) a [++ tx] = q [I]; else B [++ ty] = q [I]; solve (a, tx ), solve (B, ty);} bool bigger1 (int I, int j) {return ll [I]> ll [j];} bool bigger2 (int I, int j) {return rr [I] <rr [j];} int main () {freopen ("aut. in "," r ", stdin); freopen (" aut. out "," w ", stdout); scanf (" % d ", & n, & m); int I; www.2cto.com FOR (I, 1, m) scanf ("% d", ll + I, rr + I), p [I] = I; ll [0] = oo, rr [0] =-oo; root = build (1, n); sort (P + 1, p + m + 1, bigger1); FOR (I, 1, m) insertx (p [I]); sort (p + 1, p + m + 1, bigger2); FOR (I, 1, m) inserty (p [I]); FOR (I, 1, m) if (! C [I]) bfs (I); FOR (I, 1, m) puts (c [I] <0? "N": "S"); return 0 ;}