CodeForces 1619D New Year's Problem
1.CodeForces 1992E Novice's Mistake2.CodeForces 1935A Entertainment in MAC3.CodeForces 1992C Gorilla and Permutation4.CodeForces 1992D Test of Love5.CodeForces 1992B Angry Monk6.CodeForces 1992A Only Pluses7.CodeForces 1983B Corner Twist8.CodeForces 1983A Array Divisibility9.CodeForces 1983C Have Your Cake and Eat It Too10.CodeForces 1883B Chemistry11.CodeForces 1883A Morning12.CodeForces 1883C Raspberries13.CodeForces 1883D In Love14.CodeForces 1883E Look Back15.CodeForces 1883F You Are So Beautiful16.CodeForces 1883G1 Dances (Easy version)17.CodeForces 908B New Year and Buggy Bot18.CodeForces 908C New Year and Curling19.CodeForces 1132B Discounts
20.CodeForces 1619D New Year's Problem
21.Codeforces Round 964 (Div. 4)题目链接:CodeForces 1619D【New Year's Problem】
思路
可以因为最多只能逛n - 1个商店,当n - 1大于等于m的时候,所有朋友都能取最大值,否则至少有两个人要选择相同的商店,所以依次枚举两个人选择同一个商店,其他人选择喜悦值最大的商店。
代码
#include <cstddef> #include <cstring> #include <iostream> #include <algorithm> #include <vector> using namespace std; #define ll long long const int N = 2e5 + 10; ll a[N], num[N], maxn[N]; void solve() { memset(num, 0, sizeof num); memset(maxn, 0, sizeof maxn); // n is the number of friend int n, m; cin >> m >> n; vector<vector<ll>> ve(m + 1, vector<ll>(n + 1)); for (int i = 1; i <= m; i++) { for (int j = 1; j <= n; j++) { cin >> ve[i][j]; maxn[j] = max(maxn[j], ve[i][j]); } } ll minn = 1e9; for (int i = 1; i <= n; i++) { minn = min(minn, maxn[i]); } ll res = 0; if (n - 1 >= m) { res = minn; } else { for (int i = 1; i <= m; i++) { for (int j = 1; j <= n; j++) { for (int k = j + 1; k <= n; k++) { // 选两个人在同一个商店买礼物 res = max(res, min(minn, min(ve[i][j], ve[i][k]))); } } } } cout << res << endl; } int main() { int t; cin >> t; while (t--) { solve(); } return 0; }
合集:
Codeforces
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