Section 8.离散化（整数有序）

一、离散化的含义：

注意：

1. a[]中可能有重复元素 ，需要去重
2. 如何算出x离散化后的值：二分

二、离散化的过程

vector<int> a; //存储所有待离散化的值
sort(a.begin(), a.end()); //排序
a.erase(unique(a.begin(), a.end()), a.end());   //去重

// 二分求出x对应的离散化的值
int find(int x) // 找到第一个大于等于x的位置
{
    int l = 0, r = a.size() - 1;
    while (l < r)
    {
        int mid = l + r >> 1;
        if (alls[mid] >= x) 
            r = mid;
        else 
            l = mid + 1;
    }
    return r + 1; // 映射到1, 2, ...n;如果是返回r，则为映射到0,1,2，...n
}

三、配套练习

AcWing802https://www.acwing.com/problem/content/804/

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef pair<int, int> PII;
const int N = 300010;
int n, m;
int a[N], s[N];
vector<int> alls;
vector<PII> add, query;
int find(int x)
{
    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;
}
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);
    }
    // 去重
    sort(alls.begin(), alls.end());
    alls.erase(unique(alls.begin(), alls.end()), alls.end());
    // 处理插入
    for (auto item : add)
    {
        int x = find(item.first);
        a[x] += item.second;
    }
    // 预处理前缀和
    for (int i = 1; i <= alls.size(); i ++ ) s[i] = s[i - 1] + a[i];
    // 处理询问
    for (auto item : query)
    {
        int l = find(item.first), r = find(item.second);
        cout << s[r] - s[l - 1] << endl;
    }
    return 0;
}
//链接：https://www.acwing.com/activity/content/code/content/40105/