2038: [2009 countries Training team] small Z socks (hose) MO Team Algorithm

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First, the query is processed into Manhattan's smallest spanning tree.

Then you can run violently on the tree.

Can use MO team situation should be for inquiry [L,r], [l ', R '] cost must be ABS (L-L ') + ABS (R-R ')

#include <stdio.h> #include <iostream> #include <algorithm> #include <sstream> #include < stdlib.h> #include <string.h> #include <limits.h> #include <vector> #include <string># Include <time.h> #include <math.h> #include <iomanip> #include <queue> #include <stack># Include <set> #include <map>const int inf = 1e9;const double eps = 1e-8;const double pi = ACOs ( -1.0); Template &  Lt;class t>inline BOOL Rd (T &ret) {char c; int sgn;if (c = GetChar (), c = = EOF) return 0;while (c! = '-' && (c< ' 0 ' | | c> ' 9 ')) c = GetChar (); sgn = (c = = '-')? -1:1;ret = (c = = '-')? 0: (C-' 0 '); while (c = GetChar (), C >= ' 0 ' &&c <= ' 9 ') ret = ret * + (C-' 0 '); ret *= Sgn;return 1;} Template <class t>inline void pt (T x) {if (x <0) {Putchar ('-'); x = x;} if (x>9) pt (X/10);p Utchar (x% 10 + ' 0 ');} using namespace Std;const int N = 1e5 + 10;typedef long long ll;ll gcd (ll x, ll y) {if (X &Gt Y) Swap (x, y), while (x) y%= x, swap (x, y), return y; Vector<int>g[n];class mst{struct Edge{int from, to, dis; Edge (int _from = 0, int _to = 0, int _dis = 0): from (_from), to (_to), dis (_dis) {}bool operator < (const Edge &AMP;X) CO Nst{return dis < X.dis;}} Edge[n << 3];int F[n], tot;int find (int x) {return x = = F[x]? x:f[x] = Find (F[x]);} BOOL Union (int x, int y) {x = find (x); y = find (y); if (x = = y) return false;if (x > Y) Swap (x, y); F[x] = Y;return true;} Public:void init (int n) {for (int i = 0; I <= N; i++) F[i] = I;tot = 0;} void Add (int u, int v, int dis) {edge[tot++] = Edge (U, v, dis);} ll work () {//Calculate minimum spanning tree, return cost sort (edge, Edge + tot); ll costs = 0;for (int i = 0; i < tot; i++) if (Union (Edge[i].from, Edge[i] . To)) {cost + = Edge[i].dis; G[edge[i].from].push_back (edge[i].to); G[edge[i].to].push_back (Edge[i].from);} return cost;}} Mst;struct point{//Two-dimensional planar dots int x, y, Id;bool operator < (const point&a) Const{return x = = a.x? y < a.y:x < a . x;}} P[n];bool cmp_id (cOnst point&a, const point&b) {return a.id < b.id;} Class bit{//tree array int c[n], id[n], maxn;int lowbit (int x) {return x&-x;} Public:void init (int n) {MAXN = n + 10;fill (c, C + MAXN + 1, INF), Fill (ID, id + MAXN + 1,-1);} void Updata (int x, int val, int _id) {while (x) {if (Val < c[x]) {c[x] = val; id[x] = _id;} X-= Lowbit (x);}} int query (int x) {int val = inf, _id = -1;while (x <= maxn) {if (val > C[x]) {val = c[x]; _id = id[x];} x + = Lowbit (x);} return _id;}} tree;inline BOOL CMP (int *x, int *y) {return *x < *y;} Class Manhattan_mst{int A[n], B[n];p ublic:ll work (int L, int. r) {Mst.init (R); for (int dir = 1; dir <= 4; dir++) {if (dir% 2==0) for (int i = l; I <= R; i++) swap (p[i].x, p[i].y) and else if (dir = = 3) for (int i = l; I <= R; i++) P[i].y =-p[i].y ; sort (P + L, p + R + 1); for (int i = l; I <= R; i++) A[i] = b[i] = p[i].y-p[i].x; Discretization Sort (b + 1, B + n + 1), int sz = unique (b + 1, B + n + 1)-b-1;//initializes the inverse Tree array tree.init (SZ); for (int i = r; I >= l; i --) {int pos = Lower_bound (b + 1, B + sz + 1, a[i])-B;int id = tree.query (POS); if (id! =-1) mst.add (p[i].id, P[id].id, ABS (p[i].x-p[ id].x) + ABS (P[I].Y-P[ID].Y)) Tree.updata (POS, p[i].x + p[i].y, i);}} for (int i = l; I <= R; i++) P[i].y =-p[i].y;return mst.work ();}}  M_mst;ll Up[n], Now;int l[n], r[n];int N, Query, Col[n], siz[n];void Add (int x, int y) {for (int i = x; i <= y; i++) {Now + = siz[col[i]];siz[col[i]]++;}} void del (int x, int y) {for (int i = x; i <= y; i++) {now-= Siz[col[i]]-1;siz[col[i]]--;}} void Dfs (int u, int fa) {if (FA = =-1) Add (L[u], r[u]), Else{if (L[u] < L[FA]) Add (L[u], L[fa]-1); if (R[u] > R[FA]) A DD (R[FA] + 1, r[u]), if (L[u] > L[fa]) del (L[FA], L[u]-1), if (R[u] < R[FA]) del (R[u] + 1, R[FA]);} Up[u] = now;for (int i = 0; i < g[u].size (); i++) if (g[u][i]! = FA) Dfs (g[u][i), u); if (FA! =-1) {if (L[u] < L[FA]) d El (L[u], L[fa]-1), if (R[u] > R[fa]) del (R[FA] + 1, r[u]), if (L[u] > L[fa]) Add (L[FA], L[u]-1); if (R[u] < R[FA ]) Add (R[u] + 1, r[fA]);}} int main () {while (CIN >> n >> query) {for (int i = 1; I <= n; i++) Rd (Col[i]), for (int i = 1; I <= query; i++) {rd (L[i]), RD (R[i]);p [i].x = L[i]; p[i].y = R[i]; p[i].id = i;} for (int i = 1; I <= query; i++) g[i].clear (); m_mst.work (1, query); now = 0;memset (siz, 0, sizeof siz);d FS (1,-1); t i = 1; I <= query; i++) {ll down = (LL) (R[i]-l[i] + 1) * (R[i] – l[i])/2;ll g = gcd (Up[i], down);p t (Up[i]/g), Putchar ('/'); Pt (down/g); Putchar (' \ n ');}} return 0;}

2038: [2009 countries Training team] small Z socks (hose) MO Team Algorithm