https://atcoder.jp/contests/abc238/tasks/abc238_e
#include<iostream>
#include<cstring>
#include<cmath> 
#include<map> 
#include<algorithm>
using namespace std;

const int N = 1e6 + 10;

int n, q;
int p[N];
int find (int tar) {
    if (p[tar] != tar) return p[tar] = find(p[tar]);
    return p[tar];
}  
void solve() {
    cin >> n >> q;
    for (int i = 0; i <= n; i ++) p[i] = i;
    while ( q--) {
        int l, r;
        cin >> l >> r;
        if (l > r) swap(l, r);
        int fl = find(l - 1), fr = find(r);
        if (fl != fr) {
            p[fr] = fl;
        }
    }
    int p0 = find(0), pn = find(n);
    if (p0== pn) puts("Yes");
    else puts("No"); 
}
int main () {
    int t;
    t =1;
    while (t --) solve();
    return 0;
}

前缀和表示 a[l] + ... + a[r] = b[r] - b[l - 1]; 如果b[0]能求b[n]当且仅当0和n有边，这里可以用并查集来实现