每日一题-离散化

Interval Sum

vector<int> all;
vector<pair<int, int> > add, query;

const int N = 3e5 + 5; // Attention! When worst case that l, r, x all different, we need 3e5 to store though n, m are no bigger than 1e5.
int a[N], sum[N];

int find(int x) {
	int l = 0, r = all.size() - 1;
	
	while (l < r) {
		int mid = (l + r) >> 1;
		if (all[mid] >= x) {
			r = mid;
		} else {
			l = mid + 1;
		}
	}
	
	return l + 1;
}

int main() {
	IOS;
	
	int n, m;
	
	cin >> n >> m;
	
	for (int i = 0; i < n; ++i) {
		int x, c;
		cin >> x >> c;
		
		add.push_back(make_pair(x, c));
		
		all.push_back(x);
	}
	
	for (int i = 0; i < m; ++i) {
		int l, r;
		cin >> l >> r;
		
		query.push_back(make_pair(l, r));
		
		all.push_back(l);
		all.push_back(r);
	}
	sort(all.begin(), all.end());
	
	all.erase(unique(all.begin(), all.end()), all.end());
	
	for (int i = 0; i < n; ++i) {
		int x = find(add[i].first);
		a[x] += add[i].second;
	}

	for (int i = 1; i <= (int)all.size(); ++i) {
		sum[i] = sum[i - 1] + a[i];
	}
	
	for (auto i: query) {
		int l = find(i.first), r = find(i.second);
		
		cout << sum[r] - sum[l - 1] << '\n';
	}
    return 0;
}

Description

Given an infinite number axis with all points have value 0, oprate n times, each add point \(p_i\) by \(x_i\). Then m queries, each ask the interval sum of interval \([l, r]\). Interval sum is the sum of point value from \(l\) to \(r\).

Discretization

Establishing the mapping relationship between a number sequence and natural numbers is what discretization does. Usually we can sort the value array, then erase the duplicates, then we use \(Binary-Search\) to find the new index. After discretization, we can implement \(pre-sum\) algorithm to solve the problem.