按左端點排序，然後維護右端點最大值，貪心的思想。。。明明線掃一遍O（n）就好……然額……還是想得有些復雜

51Nod 1091 線段重疊 傳送門

#include<iostream>   
#include<algorithm>
#include<vector> 
#include<string.h>
#include<stack>
using namespace std;
typedef long long ll;
const int MAX = 5e4 + 5;
int n;
struct line {
    int l, r;
}a[MAX];
 
bool cmp(line a, line b)
{
    if (a.l == b.l)
        return a.r > b.r;
    return a.l < b.l;
}
int main()
{
    ios::sync_with_stdio(false);
    while (cin >> n)
    {
        for (int i = 0; i < n; i++)
            cin >> a[i].l >> a[i].r;
        sort(a, a + n, cmp);
        
 
int maxx=0,e=a[0].r;
        for (int i = 1; i < n; i++)
        {
            maxx = max(maxx, min(e, a[i].r) - a[i].l); //當前和之前相比較小的右端點-當前左端點
            e = max(e, a[i].r); //維護右端點最大值
        }
        cout << maxx << endl;
    }
    return 0;
}

51Nod 1091 線段重疊 貪心 區間重疊