#1167429 (C++23) No.3560 Giant Salamander

• 問題: No.3560 Giant Salamander
• ユーザー: ponjuice
• 提出日時: 2026-05-29 20:11:46
• 言語: C++23 (gcc 15.2.0 + boost 1.89.0)
• 結果: AC
• 実行時間: 305 ms / 2,000 ms
• コード長: 3,616 bytes
• コンパイル時間: 2,193 ms
• コンパイル使用メモリ: 345,032 KB
• 実行使用メモリ: 34,228 KB
• 最終ジャッジ日時: 2026-05-29 20:11:54
• 合計ジャッジ時間: 7,823 ms
• ジャッジサーバーID （参考情報）: judge4_1 / judge1_0

サブタスク	配点	結果
部分点1	5 %	AC * 5
部分点2	20 %	AC * 6
部分点3	15 %	AC * 11
部分点4	40 %	AC * 22
満点	20 %	AC * 35
合計	100 点

ソースコード

#include <bits/stdc++.h>
using namespace std;
#include<atcoder/lazysegtree>

//高速化 
struct ponjuice{ponjuice(){cin.tie(0);ios::sync_with_stdio(0);cout<<fixed<<setprecision(20);}}PonJuice;
//#define endl '\n' //インタラクティブ問題の時は消す

//型
using ll = long long;
using ld = long double;
template<class T>using vc = vector<T>; template<class T>using vvc = vc<vc<T>>; template<class T>using vvvc = vvc<vc<T>>;
using vi = vc<int>;  using vvi = vvc<int>;  using vvvi = vvvc<int>;
using vl = vc<ll>;   using vvl = vvc<ll>;   using vvvl = vvvc<ll>;
using pi = pair<int, int>;  using pl = pair<ll, ll>;
using ull = unsigned ll;
template<class T>using priq = priority_queue<T>;
template<class T>using priqg = priority_queue<T, vc<T>, greater<T>>;

// for文
#define overload4(a, b, c, d, e, ...) e
#define rep1(n)             for(ll i = 0; i < n; i++)
#define rep2(i, n)          for(ll i = 0; i < n; i++)
#define rep3(i, a, b)       for(ll i = a; i < b; i++)
#define rep4(i, a, b, step) for(ll i = a; i < b; i+= step)
#define rep(...) overload4(__VA_ARGS__, rep4, rep3, rep2, rep1)(__VA_ARGS__)
#define per1(n)             for(ll i = n-1; i >= 0; i--)
#define per2(i, n)          for(ll i = n-1; i >= 0; i--)
#define per3(i, a, b)       for(ll i = b-1; i >= a; i--)
#define per4(i, a, b, step) for(ll i = b-1; i >= a; i-= step)
#define per(...) overload4(__VA_ARGS__, per4, per3, per2, per1)(__VA_ARGS__)

//関数
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define si(x) (ll)(x).size()
template<class S, class T>inline bool chmax(S& a, T b){return a < b && ( a = b , true);}
template<class S, class T>inline bool chmin(S& a, T b){return a > b && ( a = b , true);}

inline void yes(){cout << "Yes\n";}
inline void no(){cout << "No\n";}
inline void yesno(bool y = true){if(y)yes();else no();}

//定数
constexpr ll mod = 998244353;
constexpr ll minf=-(1<<29);
constexpr ll inf=(1<<29);
constexpr ll MINF=-(1LL<<60);
constexpr ll INF=(1LL<<60);
const int dx[4] ={-1, 0, 1, 0};
const int dy[4] ={ 0, 1, 0,-1};
const int dx8[8] ={-1,-1,-1, 0, 1, 1, 1, 0};
const int dy8[8] ={-1, 0, 1, 1, 1, 0,-1,-1};

void solve();
int main() {
	int t = 1;
    cin >>t;
    while(t--)solve();
}

using S=ll;
using F=ll;
const S Inf = 8e18;
const F ID = 0;
S op(S a, S b){ return min(a,b); }
S e(){ return Inf; }
S mapping(F f, S x){ return f+x; }
F composition(F f, F g){ return f+g; }
F id(){ return ID; }

void solve(){
    ll n,m;
    cin >> n >> m;
    vector<ll> a(m), b(m);
    rep(i,0,m) cin >> a[i] >> b[i];
    if(m == 1) {
        cout << "Yes" << endl;
        return;
    }

    atcoder::lazy_segtree<F,op,e,S,mapping,composition,id> seg(m*2);
    atcoder::lazy_segtree<F,op,e,S,mapping,composition,id> seg2(m*2);
    seg.set(0, n+a[1] - a[m-1] - 1 - b[0]);
    seg2.set(0, a[1] - a[0]-1 - seg.get(0));
    rep(i,1,m*2) {
        ll add = (a[(i+1)%m] - a[i%m]);
        if(add < 0) add += n;
        seg.set(i, seg.get(i-1) + add - b[i%m]);
        seg2.set(i, (a[(i+1)%m] - a[i%m]>=0?a[(i+1)%m] - a[i%m]:a[(i+1)%m] - a[i%m]+n)-1 - seg.get(i));
    }
    
    // rep(i,0,m*2) cout << seg.get(i) << " "; cout << endl;
    // rep(i,0,m*2) cout << seg2.get(i) << " "; cout << endl;

    bool yes = false;
    rep(i,0,m) {
        ll diff = a[(i+1)%m] - a[(i+m-1)%m] - 1;
        if(diff < 0) diff += n;
        diff -= b[i];
        diff = diff - seg.get(i);
        seg.apply(i, m*2, diff);
        seg2.apply(i, m*2, -diff);
        if(seg.prod(i, i+m) >= 0 && seg2.prod(i,i+m) >= 0) {
            yes = true;
        }
    }

    cout << (yes?"Yes":"No") << endl;
}