常见问题分类：

1）区间问题

2）Huffman

3）排序不等式

4）绝对值不等式

5）推公式

 一、区间问题

板子：

①区间选点问题

#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1e5 + 10;
struct Range{
    int l, r;
    bool operator< (const Range& b) const{
        return r < b.r;
    }
}range[N];
int main(){
    int n;  cin >> n;
    int res = 0;
    for(int i = 0; i < n; i ++){
        int l, r;   cin >> l >> r;
        range[i] = {l , r};
    }
    sort(range, range + n);
    int p = -2e9;
    for(int i = 0; i < n; i ++)
        if(p < range[i].l){
            res ++;
            p = range[i].r;
        }
    cout << res;
    return 0;
}

②区间分组

#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
const int N = 1e5 +10;
struct Range{
    int l, r;
    bool operator<(const Range& b)const {
        return l < b.l;
    }
}range[N];

int n;

int main(){
    cin >> n;  
    priority_queue<int, vector<int>, greater<int>>  heap;   // 小根堆
    for(int i = 0; i < n; i ++){
        int l, r;   cin >> l >> r;
        range[i] = {l, r};
    }
    sort(range, range + n);
    for(int i = 0; i < n; i ++){
        if(heap.empty() || heap.top() >= range[i].l){
            heap.push(range[i].r);
        }else{  // 直接接到最先结束的队伍后面
            heap.pop();
            heap.push(range[i].r);
        }
    }
    cout << heap.size();
    return 0;
}

③区间覆盖问题

#include<iostream>
#include<algorithm>
using namespace std;
const int N = 1e5 + 10;
struct Range{
    int l, r;
    bool operator<(const Range& b) const {
        return l < b.l;
    }
}range[N];

int n, res;
int st, ed;

int main(){
    cin >> st >> ed;
    cin >> n;
    for(int i = 0; i < n; i ++){
        int l, r;   cin >> l >> r;
        range[i] = {l, r};
    }
    sort(range, range + n);
    
    bool success = false;
    for(int i = 0; i < n; i ++){
        int j = i, r = -2e9;  // j从当前下标开始找,贪心到r最远能到的区间
        while(j < n && range[j].l <= st){   // 能接得上
            r = max(r, range[j].r);         // 求出在接得上的前提下能够到达的最远距离
            j ++;
        }
        
        if(r < st){                         //在接得上的前提下能够到达的最远距离小于上一个线段的末尾
            res = -1;   break;              // 则说明无法覆盖
        }
        res ++;
        if(r >= ed){
            success = true;
            break;
        }
        i = j - 1;
        st = r;
    }
    if(!success)    res = -1;
    cout << res;
    return 0;
}

 Leetcode 3111.覆盖所有点的最少矩形数目

class Solution {
public: 
    int minRectanglesToCoverPoints(vector<vector<int>>& points, int w) {
        int n = points.size();
        long long res = 0;
        sort(points.begin(), points.end(), [&](const vector<int>& a, const vector<int>& b) -> bool{
            return a[0] < b[0];
        });     // 按照x坐标进行排序
        int st = points[0][0];
        int cur_l = st, cur_r = cur_l + w;  res ++;
        for (int i = 0; i < n; i ++){
            if (points[i][0] <= cur_r)  continue;
            // 否则需要新开一个区间
            cur_r = points[i][0] + w;   res ++;
        }
        return res;
    }
};