离散化模板

https://www.acwing.com/problem/content/description/804/

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<vector>

using namespace std;

typedef pair<int,int>PII;

const int N = 300010;//n+2m个坐标数量,即最多用3*1e5次方

vector<PII>add,query;//存储操作
vector<int>alls;//离散化后的下标
int a[N];//存储值
int s[N];
int n,m;

int find(int x)//寻找下标离散化后的结果
{
	int l=0,r=alls.size()-1;
	while(l<r)
	{
		int mid=l+r>>1;
		if(alls[mid]>=x)r=mid;
		else l=mid+1;
	}
	return r+1;//下标从1开始
}

int main()
{
	cin >> n >> m;
	for(int i=0;i<n;i++)
	{
		int x,c;
		cin >> x >> c;
		add.push_back({x,c});
		alls.push_back(x);
	}
	for(int i=0;i<m;i++)
	{
		int l,r;
		cin >> l >> r;
		query.push_back({l,r});
		alls.push_back(l);
		alls.push_back(r);
	}
	sort(alls.begin(),alls.end());
	alls.erase(unique(alls.begin(),alls.end()),alls.end());//unique删除重复元素,返回剩下元素的结尾位置
	//这样便实现了离散化,映射成功
	for(auto item:add)
	{
		int x=find(item.first);
		a[x]+=item.second;
	}
	for(int i=1;i<=alls.size();i++)s[i]=s[i-1]+a[i];
	for(auto item:query)
	{
		int l=find(item.first),r=find(item.second);
		cout << s[l]-s[r-1] << endl;
	}
	
}