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Show all threads Hide all threads Show all messages Hide all messages | solution C++ | Der1cle | 2005. Taxi for Programmers | 17 Jan 2020 16:48 | 1 | #include <iostream> #include <string> #include <vector> #include <set> #include <map> #include <cmath> #include <deque> #include <queue> #include <algorithm> #include <iterator> #include <bitset> #include <iomanip> #include <functional> #include <list> #include <cctype> #include <array> #include <algorithm> using namespace std; int main() { int n; n = 5; vector<vector<int>> vec(n); for (int i = 0; i < n; ++i) { vec[i].resize(n); for (int j = 0; j < n; j++) { cin >> vec[i][j]; } } cout << min(min(min(vec[0][0] + vec[1][0] + vec[2][1] + vec[2][3] + vec[3][4], vec[0][0] + vec[3][0] + vec[2][3] + vec[1][2] + vec[1][4]), vec[0][0] + vec[0][2] + vec[1][2] + vec[1][3] + vec[3][4]), vec[0][0] + vec[0][2] + vec[2][3] + vec[1][3] + vec[1][4]) << endl; if (min(min(min(vec[0][0] + vec[1][0] + vec[2][1] + vec[2][3] + vec[3][4], vec[0][0] + vec[3][0] + vec[2][3] + vec[1][2] + vec[1][4]), vec[0][0] + vec[0][2] + vec[1][2] + vec[1][3] + vec[3][4]), vec[0][0] + vec[0][2] + vec[2][3] + vec[1][3] + vec[1][4]) == vec[0][0] + vec[1][0] + vec[2][1] + vec[2][3] + vec[3][4]) { cout << "1 2 3 4 5"; } else if(min(min(min(vec[0][0] + vec[1][0] + vec[2][1] + vec[2][3] + vec[3][4], vec[0][0] + vec[3][0] + vec[2][3] + vec[1][2] + vec[1][4]), vec[0][0] + vec[0][2] + vec[1][2] + vec[1][3] + vec[3][4]), vec[0][0] + vec[0][2] + vec[2][3] + vec[1][3] + vec[1][4]) == vec[0][0] + vec[3][0] + vec[2][3] + vec[1][2] + vec[1][4]) { cout << "1 4 3 2 5"; } else if (min(min(min(vec[0][0] + vec[1][0] + vec[2][1] + vec[2][3] + vec[3][4], vec[0][0] + vec[3][0] + vec[2][3] + vec[1][2] + vec[1][4]), vec[0][0] + vec[0][2] + vec[1][2] + vec[1][3] + vec[3][4]), vec[0][0] + vec[0][2] + vec[2][3] + vec[1][3] + vec[1][4]) == vec[0][0] + vec[0][2] + vec[1][2] + vec[1][3] + vec[3][4]) { cout << "1 3 2 4 5"; } else { cout << "1 3 4 2 5"; } cout << endl; } | Hint, very easy | http://www.HelloACM.com | 2005. Taxi for Programmers | 30 Oct 2016 17:49 | 3 | s1 = ['1 2 3 4 5', '1 4 3 2 5', '1 3 2 4 5', '1 3 4 2 5'] [deleted] Edited by author 24.06.2014 02:43 int order[3] = {2, 3, 4}; do { if (order[2] == 3) continue; calculate_distance(); } while (next_permutation(order, order + 3); | Условие) | Akula | 2005. Taxi for Programmers | 12 Aug 2016 23:39 | 2 | С чего это Илья не хочет быть отвезен последним?) Боится остаться с дядей Мишей наедине :D:D На перекрестке Крауля-Токарей развязка неудобная, а Илья беспокоится о тренере | Why WA #2 | Najmaddin Akhundov | 2005. Taxi for Programmers | 24 Jan 2015 10:12 | 3 | Why WA #2 Najmaddin Akhundov 20 Nov 2014 04:15 [code deleted] Edited by moderator 19.11.2019 23:01 I got it. If there are two possible answer, only 1 of them should be printed. If there are two possible answer, only 1 of them should be printed. Thanks...!!! | дайте пару вариантов тестов | Andriy | 2005. Taxi for Programmers | 8 Nov 2014 15:48 | 1 | |
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