E - Reverse the Rivers 二分与思维

原题

E. Reverse the Rivers

思路

从上往下一定是越来越大，因此分成两个去看，即对于大于和小于分开去思考

代码

#include <bits/stdc++.h>
using namespace std;

int main()
{
	int n, k, q;
	cin >> n >> k >> q;
	
	vector<vector<int>> a(k, vector<int> (n));
	
	for (int i = 0; i < n; i ++ )
	{
		for (int j = 0; j < k; j ++ )
		{
			cin >> a[j][i];
			if (i > 0)
				a[j][i] |= a[j][i - 1];
		}
	}
	
	while (q -- )
	{
		int m;
		cin >> m;
		
		int low = 0, high = n;
		while (m -- )
		{
			int r, x;
			char c;
			cin >> r >> c >> x;
			r --;
			if (c == '>')
			{
				low = max(low, int(upper_bound(a[r].begin(), a[r].end(), x) - a[r].begin()));
			}
			else
			{
				high = min(high, int(lower_bound(a[r].begin(), a[r].end(), x) - a[r].begin()));
			}
		}
		
		cout << (low < high ? low + 1: -1) << endl;
	}
}