• 方法1：使用结构体存储数据，经过排序后，依次计算每个区间的数量即可

• 方法2：数据过大，将数据离散化一下，进行差分即可

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e5 + 10, INF = 0x3f3f3f3f, MOD = 1E9 + 7;

struct Edge {
    int l, r;
} edge[N];
int n, m, b[N], d[N], idx;
int find(int x) {
    return lower_bound(b + 1, b + 1 + m, x) - b;
}
bool cmp(Edge a, Edge b) {
    return a.l < b.l;
}
void solve2() { // 结构体排序 + 模拟 
    sort(edge + 1, edge + 1 + n, cmp);
    int ed = edge[1].l, tt = 0;
    for (int i = 1; i <= n; i++) {
        if (ed < edge[i].l)
            tt += edge[i].r - edge[i].l;
        else
            tt += max(edge[i].r - ed, 0);
        ed = max(ed, edge[i].r);
    }
    cout << tt;
}
void solve1(){ // 离散化 + 二分 
	sort(b + 1, b + 1 + idx);
     m = unique(b + 1, b + 1 + idx) - b - 1;
     for (int i = 1; i <= n; i++) {
         int l = find(edge[i].l);
         int r = find(edge[i].r);
         d[l]++, d[r]--;
     }
     int tt = 0;
     for (int i = 1; i < m; i++) {
         d[i] += d[i - 1];
         if (d[i])
             tt += b[i + 1] - b[i];
     }
     cout << tt;
}
int main() {
    cin >> n;
    for (int i = 1, l, r; i <= n; i++) {
        cin >> l >> r, edge[i] = {l, r};
        b[++idx] = l, b[++idx] = r;
    }
    // solve1();
   solve2();
    return 0; 
}