Educational Codeforces Round 80 D. Minimax Problem(二分,状态压缩)
题目链接
Educational Codeforces Round 80 D. Minimax Problem
思路
看到最小值最大我们很容易想到二分,看到 m m m最大为 8 8 8我们很容易想到状态压缩。
我们使用二分答案,考虑如何去check。
在check函数里面,我们将原矩阵中大于等于 m i d mid mid的值视为 1 1 1,小于 m i d mid mid的值视为 0 0 0。按照每一行进行二进制状压,因为 2 m 2^{m} 2m最大为 256 256 256,因此我们直接使用桶来记录每一行状压的结果。
暴力枚举两种状态,如果这两种状态都出现过且 o r or or起来为 2 m − 1 2^{m}-1 2m−1就满足条件。
代码
#include <bits/stdc++.h>
using namespace std;
#define int long long
typedef pair<int, int> pii;
const int N = 3e5 + 5, M = 10;
const int mod = 998244353;
const int inf = 0x3f3f3f3f3f3f3f3f;
int n, m;
int a[N][M], st[(1ll << 9)];
int qmi(int a, int b)
{
int res = 1;
while (b)
{
if (b & 1) res = res * a;
b >>= 1;
a = a * a;
}
return res;
}
bool check(int x, int &tx, int &ty)
{
memset(st, 0, sizeof st);
for (int i = 1; i <= n; i++)
{
int res = 0;
for (int j = 1; j <= m; j++)
{
res = res * 2 + (a[i][j] >= x);
}
st[res] = i;
}
for (int i = 0; i <= qmi(2, m) - 1; i++)
{
for (int j = 0; j <= qmi(2, m) - 1; j++)
{
if (st[i] && st[j] && (i | j) == qmi(2, m) - 1)
{
tx = st[i], ty = st[j];
return true;
}
}
}
return false;
}
void solve()
{
cin >> n >> m;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
cin >> a[i][j];
}
}
int low = 0, high = 1e9;
int tx = 1, ty = 1;
while (low < high)
{
int mid = low + high + 1 >> 1;
if (check(mid, tx, ty))
{
low = mid;
}
else
{
high = mid - 1;
}
}
cout << tx << " " << ty << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int test = 1;
// cin >> test;
for (int i = 1; i <= test; i++)
{
solve();
}
return 0;
}