ZONe Energy Programming Contest C.MAD TEAM

题目链接

思维+二分
如果暴力选三个，复杂度为 C 3000 3 C_{3000}^3 C30003，显然不可取，可以二分答案 a n s ans ans，这样可以把每个属性值更新成 0 0 0 或 1 1 1 (判断 a n s ans ans 和属性值大小)而言，这样五个属性值的情况最多为 2 5 = 32 2^5=32 25=32，再从中暴力选即可，AC代码如下：

#include <bits/stdc++.h>using namespace std;typedef long long ll;
int n, flag = 0;
vector<vector<int>> a;
set<int> s;void dfs(int id, int num) {if (id == 3) {if (num == 31) {flag = 1;}return;}for (auto i:s) {dfs(id + 1, num | i);}
}bool check(int x) {s.clear();flag = 0;for (int i = 0; i < n; i++) {int num = 0;for (int j = 0; j < 5; j++) {if (a[i][j] >= x) num |= (1 << j);}s.insert(num);}dfs(0, 0);return flag;
}int main() {cin >> n;a.resize(n, vector<int>(5));for (int i = 0; i < n; i++) {for (int j = 0; j < 5; j++) {cin >> a[i][j];}}int l = 1, r = 1e9;while (l <= r) {int mid = l + r >> 1;if (check(mid)) l = mid + 1;else r = mid - 1;}cout << r << endl;
}

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