两侧覆盖问题解

线段树 + 生成树染色，标准O(nlogn)，再也不用担心会被卡了。。。

#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(a) (a).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 = (a).begin(); i != (a).end(); ++i)typedef unsigned int uint;typedef long long int64;typedef unsigned long long uint64;typedef long double real;template<class T> inline bool minim(T &a, const T &b) {return b < a ? a = b, 1 : 0;}template<class T> inline bool maxim(T &a, const T &b) {return b > a ? 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 = 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 = 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 = 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%d", &n, &m);int i;FOR(i, 1, m) scanf("%d%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;}