[hihoCoder1236 Scores 2015BeijingOnline] Simple Rough sub-block + simple rough bitset

Test instructions: 50,000 5-dimensional vectors, 50,000 queries each dimension is not much more than the number of vectors that have always been, forcing online.

Idea: Finish This problem only know bitset efficiency so high, own local test 1s can operate 1010 bit,orz simple Rough

• S (i) represents the i-dimensional vector set of the current vector, the answer is count (∩s (i)), and 0≤i‹5
• Each vector can be bound with an ID (take off the mark on the line), after the binding can consider bitset. Sorted by each dimension, bitset of the ID before the current position is preprocessed at each √n location
• When querying, the maximum position is obtained for each dimension, then S (i) can be obtained with 1 preprocessing results + √n single bit insertion of up to two points. Then it's and and count.

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#include <bits/stdc++.h>using namespace std;#ifndef Online_judge #include "Local.h"#endif#define X First#define Y Second#define PB (x) push_back (x)#define MP (x, y) make_pair (x, y)#define ALL (a) (a). Begin (), (a). End ()#define MSET (A, X) memset (A, X, sizeof (a))#define MCPY (A, B) memcpy (A, B, sizeof (a))typedef Long Long ll;Template<TypeName T>BOOL Umax (T&A, Const T&b) {return b<=a?false:(A=b,true);}Template<TypeName T>BOOL Umin (T&A, Const T&b) {return b>=a?false:(A=b,true);}int K;struct Node { int a[5]; int ID; void Read() { for (int I = 0; I < 5; I ++) { scanf ("%d", a + i); } } BOOL operator < (Const Node &That ) Const { return A[k] < That.a[k]; }};Node stu[1 << -];int r[5][1 << -], id[5][1 << -];Bitset<1 << -> bs[5][1 << 8];int Main() {#ifndef Online_judge Freopen ("In.txt", "R", stdin); //freopen ("OUT.txt", "w", stdout);#endif//Online_judge int T, N, Q, m; Cin >> T; while (T --) { Cin >> N >> m; for (int I = 0; I < N; I ++) { Stu[i].read (); stu[i].id = i; } int L = sqrt (n + 0.5); for (int I = 0; I < 5; I ++) { k = i; sort (Stu, Stu + n); Bitset<1 << -> BBS; for (int J = 0; J < N; J ++) { R[i][j] = Stu[j].a[i]; Id[i][j] = stu[j].id; Bbs[stu[j].id] = 1; if (J % L == L - 1 || J == N - 1) { bs[i][j / L] = BBS; } } } int Lastans = 0; Cin >> Q; while (q --) { Node qry; Qry.read (); Bitset<1 << -> ans; for (int I = 0; I < 5; I ++) { Qry.a[i] ^= Lastans; Bitset<1 << -> buf; int POS = Upper_bound (R[i], R[i] + N, Qry.a[i]) - R[i]; if (POS / L) buf = Bs[i][pos / L - 1]; while (POS % L) buf[id[i][-- POS]] = 1; if (i) ans &= buf; Else ans = buf; } printf ("%d\ n", Lastans = Ans.count ()); } } return 0;}

[hihoCoder1236 Scores 2015BeijingOnline] Simple Rough sub-block + simple rough bitset