E. Tracking Segments

链接

https://codeforces.com/problemset/problem/1843/E

题面

思路

二分加树状数组。关键点在于看出来单点修改和区间查询，然后离线+二分：令l=1（1次操作），r=q（最多q次操作）。二分判断能不能行。
以及树状数组的板子要记得。

代码

#define _CRT_SECURE_NO_WARNINGS 
#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
#include<sstream>
#include<string>
#include<string.h>
#include<iomanip>
#include<stdlib.h>
#include<map>
#include<queue>
#include<limits.h>
#include<climits>
#include<fstream>
#include<stack>
#define IOS ios::sync_with_stdio(false), cin.tie(0) ,cout.tie(0)
using namespace std;
#define int long long
const int N = 1e5 + 10;

struct ele
{
	int begin, end;
	int length;
}eles[N];
bool cmp(ele a, ele b) { return a.length < b.length; }
int lowbit(int x) { return x & (-x); }
int a[N];
void update(int x)
{
	while (x < N)
	{
		a[x]++;
		x += lowbit(x);
	}
}
int n, m;
int sum(int x)
{
	int ans = 0;
	while (x > 0)
	{
		ans += a[x];
		x -= lowbit(x);
	}
	return ans;
}
vector<int>vt;
bool check(int nm)
{
	memset(a, 0, sizeof(a));
	for (int i = 1; i <= nm; i++)
	{
		update(vt[i]);
	}
	bool fa = false;
	for (int i = 1; i <= m and !fa; i++)
	{
		int xnoe = sum(eles[i].end) - sum(eles[i].begin - 1);
		if (xnoe>= eles[i].length / 2 + 1)fa = true;
	}
	return fa;

}


signed main()
{
	IOS;
	int t; cin >> t;
	while (t--)
	{
		vt.clear();
		vt.push_back(0);
		cin >> n >> m;
		for (int i = 1; i <= m; i++)
		{
			int l; int r;
			cin >> l >> r;
			eles[i].begin = l, eles[i].end = r;
			eles[i].length = r - l + 1;
		}
		int q; cin >> q;
		int x;
		for (int i = 0; i < q; i++) { cin >> x; vt.push_back(x); }
		int l = 1, r = q;
		while (l < r)
		{
			int m = (l + r ) / 2;
			if (check(m))r = m;
			else l = m + 1;
		}
		if (r == q and !check(r))cout << -1;
		else cout << r;
		cout << '\n';

		


	}

	return 0;
}