Submission #1227645

#TimeUsernameProblemLanguageResultExecution timeMemory
1227645madamadam3Aliens (IOI16_aliens)C++20
25 / 100
2093 ms22344 KiB
#include "aliens.h" #include <bits/stdc++.h> using namespace std; typedef long long ll; using vi = vector<int>; using vvi = vector<vi>; using vl = vector<ll>; using vvl = vector<vl>; const ll INF = 1e18; int n, m, k; vl r, c, L, R; void preprocess(int N, int M, int K, vi Row, vi Col) { vector<int> sorted_points(N); iota(sorted_points.begin(), sorted_points.end(), 0); sort(sorted_points.begin(), sorted_points.end(), [&](int a, int b) { ll La = min(Row[a], Col[a]), Ra = max(Row[a], Col[a]), Lb = min(Row[b], Col[b]), Rb = max(Row[b], Col[b]); return La == Lb ? Ra > Rb : La < Lb; }); vector<int> points; int lastR = -1; for (int i = 0; i < N; i++) { int cur = sorted_points[i]; int rcur = max(Row[cur], Col[cur]); if (rcur > lastR) { points.push_back(cur); lastR = rcur; } } n = points.size(); m = M; k = K; r.resize(n); c.resize(n); L.resize(n); R.resize(n); for (int i = 0; i < n; i++) { r[i] = Row[points[i]]; c[i] = Col[points[i]]; L[i] = min(r[i], c[i]); R[i] = max(r[i], c[i]); } } // n <= 50, m <= 100 ll sub1() { vvi used(m, vi(m, 0)); for (int i = 0; i < n; i++) { for (int x = L[i]; x <= R[i]; x++) { for (int y = L[i]; y <= R[i]; y++) { used[x][y]++; } } } ll ans = 0; for (int x = 0; x < m; x++) { for (int y = 0; y < m; y++) { if (used[x][y]) ans++; } } return ans; } // n <= 500, m <= 1000, r[i] = c[i] // essentially DP[i][k] = min cost to cover first i using k // DP[i][k] = min(DP[j[k-1] + (r[i] - r[j+1] + 1)^2) ll sub2() { vvl DP(n, vl(k+1, INF)); ll ans = INF; for (int kv = 1; kv <= k; kv++) { DP[0][kv] = 1; for (int i = 1; i < n; i++) { DP[i][kv] = DP[i-1][kv-1] + 1; for (int j = 0; j < i; j++) { ll prev = j == 0 ? 0 : DP[j-1][kv-1]; ll delta = (c[i] - c[j] + 1); delta *= delta; DP[i][kv] = min(DP[i][kv], prev + delta); } } ans = min(ans, DP[n-1][kv]); } return ans; } // n <= 500, m <= 1000 // same dp but we need to account for the squares, as well as overlap between them ll sub3() { vvl DP(n, vl(k+1, INF)); ll ans = INF; for (int kv = 1; kv <= k; kv++) { DP[0][kv] = (R[0] - L[0] + 1) * (R[0] - L[0] + 1); for (int i = 1; i < n; i++) { ll width = (R[i] - L[i] + 1), overlap_width = (R[i-1] - L[i] + 1); ll wd = width*width, owd = overlap_width > 0 ? overlap_width*overlap_width : 0; DP[i][kv] = DP[i-1][kv-1] + wd - owd; for (int j = 0; j < i; j++) { ll prev = j == 0 ? 0 : DP[j-1][kv-1]; ll delta = (R[i] - L[j] + 1); delta *= delta; ll overlap = 0; if (j > 0 && R[j-1] - L[j] + 1 > 0) { overlap = (R[j-1] - L[j] + 1); overlap *= overlap; } if (prev + delta - overlap < DP[i][kv]) { DP[i][kv] = prev + delta - overlap; } } } ans = min(ans, DP[n-1][kv]); } return ans; } // n <= 4000, m <= 1m /* let P = prev let O = max(0, R[j-1] - L[j] + 1)^2 let L = L[j]-1 let R = R[i] f(x) = (R-L)^2 + P + O = R^2 - 2LR + L^2 + P + O let C = L^2 + P + O ∴ f(x) = R^2 - 2LR + C so we can do a cht/li chao tree */ ll take_photos(int N, int M, int K, vi Row, vi Col) { preprocess(N, M, K, Row, Col); // return sub1(); // return sub2(); return sub3(); }

Compilation message (stderr)

aliens.h:1:9: warning: #pragma once in main file
    1 | #pragma once
      |         ^~~~
aliens_c.h:1:9: warning: #pragma once in main file
    1 | #pragma once
      |         ^~~~
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