제출 #770700

#제출 시각아이디문제언어결과실행 시간메모리
770700treewaveCloud Computing (CEOI18_clo)C++17
18 / 100
3088 ms262144 KiB
#include <bits/stdc++.h> using namespace std; #define int long long const int INF = 1e18; int32_t main() { ios_base::sync_with_stdio(0); cin.tie(0); int n; cin >> n; vector<array<int,3>> computers; for (int i = 0; i < n; i++){ int cores, clock, price; cin >> cores >> clock >> price; computers.push_back({clock, cores, price}); } int m; cin >> m; vector<array<int,3>> customers; for (int i = 0; i < m; i++){ int cores, clock, price; cin >> cores >> clock >> price; customers.push_back({clock, cores, price}); } sort(computers.begin(), computers.end()); sort(customers.begin(), customers.end()); vector<int> freq_computers; for (int i = 0; i < n; i++){ freq_computers.push_back(computers[i][0]); } vector<array<int, 2>> computer_ranges; //want to store for each customer adjacent pair, the range of computers with freq in [freq[i], freq[i+1]) //this naturally ignores the instances where freq[i] == freq[i+1], so we don't have to worry about overcount for (int i = 0; i < m-1; i++){ int lb = lower_bound(freq_computers.begin(), freq_computers.end(), customers[i][0]) - freq_computers.begin(); int ub = lower_bound(freq_computers.begin(), freq_computers.end(), customers[i+1][0]) - freq_computers.begin(); if (lb == n || lb > ub){ computer_ranges.push_back({-1, -1}); continue; } ub--; computer_ranges.push_back({lb, ub}); } int final_lb = lower_bound(freq_computers.begin(), freq_computers.end(), customers.back()[0]) - freq_computers.begin(); if (final_lb != n){ computer_ranges.push_back({final_lb, n-1}); } else{ computer_ranges.push_back({-1, -1}); } // for (int i = 0; i < m; i++){ // cout << computer_ranges[i][0] << " " << computer_ranges[i][1] << "\n"; // } //bounding to get subtasks int maxCores = 0; int temp = 0; for (int i = 0; i < m; i++){ temp += customers[i][1]; } maxCores = max(maxCores, temp); temp = 0; for (int i = 0; i < n; i++){ temp += computers[i][1]; } maxCores = max(maxCores, temp); maxCores++; vector<vector<int>> minCost(m, vector<int>(maxCores, INF)); vector<vector<int>> newMinCost(m, vector<int> (maxCores, INF)); for (int i = 0; i < m; i++){ int lb = computer_ranges[i][0], ub = computer_ranges[i][1]; if (lb == -1) continue; minCost[i][0] = 0; for (int j = lb; j <= ub; j++){ for (int k = computers[j][1]; k < maxCores; k++){ newMinCost[i][k] = min(minCost[i][k], minCost[i][k - computers[j][1]] + computers[j][2]); } for (int k = computers[j][1]; k < maxCores; k++){ minCost[i][k] = newMinCost[i][k]; } //suffix min int currMin = INF; for (int k = maxCores-1; k >= 0; k--){ minCost[i][k] = min(minCost[i][k], currMin); currMin = min(currMin, minCost[i][k]); } } } vector<vector<int>> dp(m+1, vector<int> (maxCores, -INF)); vector<vector<int>> newdp(m+1, vector<int>(maxCores, -INF)); //dp[i][c] is maximum profit using suffix [i,....,m-1] and having c extra cores dp[m][0] = 0; for (int i = 0; i < maxCores; i++){ dp[m][i] = max(dp[m][i], -minCost[m-1][i]); } //suffix max int currMax = -INF; for (int i = maxCores-1; i >= 0; i--){ dp[m][i] = max(dp[m][i], currMax); currMax = max(currMax, dp[m][i]); } for (int i = m-1; i >= 0; i--){ for (int c = 0; c < maxCores - customers[i][1]; c++){ dp[i][c] = max(dp[i+1][c], dp[i+1][c + customers[i][1]] + customers[i][2]); } currMax = -INF; for (int j = maxCores-1; j >= 0; j--){ dp[i][j] = max(dp[i][j], currMax); currMax = max(currMax, dp[i][j]); } //add cores from new computers if (i != 0){ for (int j = 0; j < maxCores; j++){ newdp[i][j] = -INF; } for (int j = 0; j < maxCores; j++){ for (int x = 0; x < maxCores - j; x++){ newdp[i][j+x] = max({newdp[i][j+x], dp[i][j+x], dp[i][j] - minCost[i-1][x]}); } } for (int j = 0; j < maxCores; j++){ dp[i][j] = newdp[i][j]; } } currMax = -INF; for (int j = maxCores-1; j >= 0; j--){ dp[i][j] = max(dp[i][j], currMax); currMax = max(currMax, dp[i][j]); } } int ans = -INF; for (int i = 0; i < maxCores; i++){ ans = max(ans, dp[0][i]); } cout << ans << "\n"; }
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