WHU 1581 Union of cubes

Title Link: http://acm.whu.edu.cn/land/problem/detail?problem_id=1581

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observed that the first dimension is the largest only $10$ so the first dimension can be directly enumerated $ ($) $

observed that the second dimension is the largest only $100$ so the second dimension can also continue to enumerate $ ($ cut the rectangle into segments $) $

If the last dimension is covered with a segment tree, the complexity is

$O (N^{2}kmlog (n * k ^ 2)) $ $ (n <=, K <=, M <= 1000) $

Since the time limit is only $500ms$, there are multiple sets of data that are likely to $t$ out.

Think about it, we'll find only the interval override operation without interval modification operation $ ($ equivalent to No fallback feature $) $

Then just maintain the right end of the overlay segment for each point.

And the maintenance operation is obviously to use and check the set is more convenient

Complexity reduced by one $log$ after a single group of $10^6$ multiple groups will not $t$

1#include <cstdio>2#include <cstring>3#include <cmath>4#include <algorithm>5 using namespacestd;6 inta[ One][ the][1010], fa[ One][ the][1010];7 intN, K0, M, N2, N3, sum, ans;8 intP1, Q1, R1, p2, Q2, R2;9 intFINDF (intXintYintz)Ten { One     if(Fa[x][y][z]! =z) AFA[X][Y][Z] =findf (x, Y, fa[x][y][z]); -     returnFa[x][y][z]; - } the intMain () - { -      while(SCANF ("%d%d%d", &n, &k0, &m)! =EOF) -     { +N2 = n *K0; -n3= N2 *K0; +          for(inti =1; I <= N; ++i) A              for(intj =1; J <= N2; ++j) at             { -                  for(intK =1; K <= N3; ++k) -                 { -scanf"%d", &a[i][j][k]); -FA[I][J][K] =K; -                 } inFa[i][j][n3 +1] = n3 +1; -             } tosum =0; +Ans =-1e9; -          while(m--) the         { *scanf"%d%d%d%d%d%d", &p1, &q1, &r1, &AMP;P2, &AMP;Q2, &r2); $              for(inti = p1; I <= p2; ++i)Panax Notoginseng                  for(intj = Q1; J <= Q2; ++j) -                      for(intK = R1; K <= R2; K =Fa[i][j][k]) the                         if(Findf (I, j, k) = =k) +                         { ASum + =A[i][j][k]; theFa[i][j][k] = FINDF (i, J, K +1); +                         } -Ans =max (ans, sum); $         } $printf"%d\n", ans); -     } -     return 0; the}

WHU 1581 Union of cubes