按左端点排序，然后维护右端点最大值，贪心的思想。。。明明线扫一遍O（n）就好……然额……还是想得有些复杂

#include
     
         #include
      
       #include
       
         #include
        
         #include
         
          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); //当前和之前相比较小的右端点-当前左端点            e = max(e, a[i].r); //维护右端点最大值        }        cout << maxx << endl;    }    return 0;}