codeforcesEducational Codeforces Round 80 D. Minimax Problem——二分+二进制处理
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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
#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;
}
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