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Shiroha白羽的博客

Golang 踩坑 —— interface 为参数的时候传 nil 指针 Codeforces Round 925 (Div. 3) Codeforces Round 924 (Div. 2) Codeforces Round 923 (Div. 3) Codeforces Round 922 (Div. 2) Codeforces Round 921 (Div. 2) Educational Codeforces Round 161 (Rated for Div. 2) Codeforces Round 920 (Div. 3) Codeforces Round 919 (Div. 2) Hello 2024 Good Bye 2023 Codeforces Round 918 (Div. 4) 个人备份的常用 macOS 清理命令 Codeforces Round 917 (Div. 2) Pinely Round 3 (Div. 1 + Div. 2) Educational Codeforces Round 160 (Rated for Div. 2) Codeforces Round 915 (Div. 2) Codeforces Round 914 (Div. 2) Codeforces Round 913 (Div. 3) Educational Codeforces Round 159 (Rated for Div. 2) Codeforces Round 912 (Div. 2) Codeforces Round 911 (Div. 2) CodeTON Round 7 (Div. 1 + Div. 2, Rated, Prizes!) Educational Codeforces Round 158 (Rated for Div. 2) Codeforces Round 910 (Div. 2) Codeforces Round 909 (Div. 3) Codeforces Round 908 (Div. 2) Educational Codeforces Round 157 (Rated for Div. 2) C++自定义的字面量 Codeforces Round 907 (Div. 2)
Educational Codeforces Round 80 D. Minimax Problem——二分+...
Shiroha · 2020-02-04 · via Shiroha白羽的博客

题目链接

题目大意

有n个维度为m的向量,取其中两个进行合并,合并时每个维度取两者之间的较大者,得到的新的向量中,维度值最小者最大为多少

分析

首先最需要注意的是m的取值,m最大只有8,那么我们可以二分答案,对于每一个二分值,进行下面的操作,将整个矩阵的每一个元素,如果这个元素大于二分值,则变成1,反正则变成0,把每一个向量压缩为单个二进制数,这样我们最多只会得到$2^8 = 256$种不同的二进制数,然后暴力的遍历所有可能的二进制数的组合,得到是否满足当前二分值

AC code

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#include <bits/stdc++.h>

using namespace std;

const int NUM = 3e5 + 100;

int data[NUM][10];

bool check(int value, int n, int m, pair<int, int> &ans) {
map<unsigned, int> s;
for (int i = 0; i < n; ++i) {
unsigned temp = 0;
for (int j = 0; j < m; ++j) {
temp <<= 1u;
temp |= data[i][j] > value;
}
s.insert({temp, i});
}
unsigned tar = -1u >> (sizeof(int) * 8 - m);
for (auto iter1 = s.begin(); iter1 != s.end(); ++iter1) {
for (auto iter2 = iter1; iter2 != s.end(); ++iter2) {
if ((iter1->first | iter2->first) == tar) {
ans.first = iter1->second;
ans.second = iter2->second;
return true;
}
}
}
return false;
}

void solve() {
int n, m;
cin >> n >> m;
int l = INT_MAX, r = 0;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
cin >> data[i][j];
l = min(l, data[i][j]);
r = max(r, data[i][j]);
}
}
int mid, cnt = r - l;
pair<int, int> ans;
while (cnt > 0) {
int step = cnt / 2;
mid = l + step;
if (check(mid, n, m, ans)) {
l = mid + 1;
cnt -= step + 1;
} else
cnt /= 2;
}
cout << ans.first + 1 << " " << ans.second + 1 << endl;
}

signed main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
#ifdef ACM_LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
long long test_index_for_debug = 1;
char acm_local_for_debug;
while (cin >> acm_local_for_debug) {
cin.putback(acm_local_for_debug);
if (test_index_for_debug > 20) {
throw runtime_error("Check the stdin!!!");
}
auto start_clock_for_debug = clock();
solve();
auto end_clock_for_debug = clock();
cout << "Test " << test_index_for_debug << " successful" << endl;
cerr << "Test " << test_index_for_debug++ << " Run Time: "
<< double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
cout << "--------------------------------------------------" << endl;
}
#else
solve();
#endif
return 0;
}