結果

問題 No.3265 地元に帰れば天才扱い！
コンテスト
ユーザー AD010
提出日時 2025-09-06 15:45:56
言語 C++17(clang)
(17.0.6 + boost 1.87.0)
結果
AC  
実行時間 2,329 ms / 2,500 ms
コード長 11,263 bytes
コンパイル時間 5,647 ms
コンパイル使用メモリ 175,076 KB
実行使用メモリ 38,184 KB
最終ジャッジ日時 2025-09-06 15:46:55
合計ジャッジ時間 57,050 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 4
other AC * 21
権限があれば一括ダウンロードができます

ソースコード

diff # 

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

#define ll long long
#define ull unsigned long long
#define ld  long double

using LL = long long; using ULL = unsigned long long;
using VI = vector<int>; using VVI = vector<VI>; using VVVI = vector<VVI>;
using VL = vector<LL>; using VVL = vector<VL>; using VVVL = vector<VVL>;
using VB = vector<bool>; using VVB = vector<VB>; using VVVB = vector<VVB>;
using VD = vector<double>; using VVD = vector<VD>; using VVVD = vector<VVD>;
using VC = vector<char>; using VS = vector<string>; using VVC = vector<VC>;
using PII = pair<int,int>; using PLL = pair<LL,LL>; using PDD = pair<double,double>; using PIL = pair<int,LL>;
using MII = map<int,int>; using MLL = map<LL,LL>;
using SI = set<int>; using SL = set<LL>;
using MSI = multiset<int>; using MSL = multiset<LL>;
template<class T> using MAXPQ = priority_queue<T>;
template<class T> using MINPQ = priority_queue< T, vector<T>, greater<T> >;

const ll MOD = 1000000007;
const ll MOD2 = 998244353;
const ll INF = 1LL << 60;
#define PI  3.14159265358979323846

#define FOR(i, a, b) for(int i = (a); i < (b); ++i)
#define REP(i, n) FOR(i, 0, n)
#define EACH(e, v) for(auto &e : v)
#define RITR(it, v) for(auto it = (v).rbegin(); it != (v).rend(); ++it)
#define ALL(v) v.begin(),v.end()

vector<ll> x8={1,1,1,0,0,-1,-1,-1},y8={1,0,-1,1,-1,1,0,-1};
int dx4[4]={1,-1,0,0}, dy4[4]={0,0,1,-1};

/*
memo
-uf,RMQ(segtree),BIT,BIT2,SegTree,SegTreeLazy
-isprime,Eratosthenes,gcdlcm,factorize,divisors,modpow,moddiv
nCr(+modnCr,inverse,extend_euclid.powmod),tobaseB,tobase10
-dijkstra,Floyd,bellmanford,sccd,topological,treediamiter
-compress1,compress2,rotate90

-co,ci,fo1,fo2,fo3,fo4
-bitsearch,binaryserach
-bfs
-SegTreedec,SegTreeLazydec
*/

template <typename X, typename M>
struct LazySegTree{
    using FX = function<X(X,X)>;
    using FA = function<X(M,X)>;
    using FM = function<M(M,M)>;

    long long n, N, log;
    vector<X> dat;
    vector<M> lazy;
    FX fx;
    FA fa;
    FM fm;
    X ex;
    M em;

    LazySegTree(long long _n, FX _fx, FA _fa, FM _fm, X _ex, M _em){
        init(_n,_fx,_fa,_fm,_ex,_em);
    }

    void init(long long _n, FX _fx, FA _fa, FM _fm, X _ex, M _em){
        N = _n, fx = _fx, fa = _fa, fm = _fm, ex = _ex, em = _em;
        long long x = 1;
        log = 0;
        while(_n > x) x *= 2, log++;
        n = x;
        dat.assign(n*2,ex);
        lazy.assign(n*2,em);
    }

    void pull_dat(long long k){
        dat[k] = fx(dat[2*k],dat[2*k+1]);
    }

    void apply_lazy(long long k, M f){
        dat[k] = fa(f, dat[k]);
        if(k < n) lazy[k] = fm(f, lazy[k]);
    }

    void push_lazy(long long k){
        if(lazy[k] == em) return;
        apply_lazy(2*k, lazy[k]);
        apply_lazy(2*k + 1, lazy[k]);
        lazy[k] = em;
    }

    void pull_dat_deep(long long k){
        for(int h = 1; h <= log; h++) pull_dat(k >> h);
    }

    void push_lazy_deep(long long k){
        for(int h = log; h >= 1; h--) push_lazy(k >> h);
    }

    //i番目の要素をxにアップデートする(0-indexed),O(logN)
    void set(long long i, X x){
        i += n;
        push_lazy_deep(i);
        dat[i] = x;
        pull_dat_deep(i);
    }

    //i番目の要素にアクセス(0-indexed),O(logN);
    X get(long long i){
        i += n;
        push_lazy_deep(i);
        return dat[i];
    }
    X operator[](long long i){return get(i);};

    //i番目の要素にfaを作用させる
    void apply(long long i, M f){
        i += n;
        push_lazy_deep(i);
        dat[i] = fa(f, dat[i]);
        pull_dat_deep(i);
    }

    //[l,r)までfaを作用させる
    void apply(long long l, long long r, M f){
        if(l == r) return;
        l += n, r += n;
        for(int h = log; h >= 1; h--){
            if(((l >> h) << h) != l) push_lazy(l >> h);
            if(((r >> h) << h) != r) push_lazy((r-1) >> h);
        }
        long long L = l, R = r;
        for(;l < r; l >>= 1, r >>= 1){
            if(l & 1) apply_lazy(l++,f);
            if(r & 1) apply_lazy(--r,f);
        }
        l = L, r = R;
        for(int h = 1; h <= log; h++){
            if(((l >> h) << h) != l) pull_dat(l >> h);
            if(((r >> h) << h) != r) pull_dat((r-1) >> h);
        }
    }
    //[l,r)で二項演算を作用した結果(0-indexed),O(logN)
    X prod(long long l, long long r){
        if(l == r) return ex;
        l += n, r += n;
        for(int h = log; h >= 1; h--){
            if(((l >> h) << h) != l) push_lazy(l >> h);
            if(((r >> h) << h) != r) push_lazy(r >> h);
        }
        X vleft = ex, vright = ex;
        for(; l < r; l >>= 1, r >>= 1){
            if(l & 1) vleft = fx(vleft,dat[l++]);
            if(r & 1) vright = fx(dat[--r],vright);
        }
        return fx(vleft,vright);
    }
    X all_prod(){return dat[1];};

    //x=prod(l,r),f(x)=trueとなる最大のrを求める(f(ex)=true, 0-indexed),O(logN)
    long long max_right(function<bool(X)> f, long long l = 0){
        if(l == N) return N;
        l += n;
        push_lazy_deep(l);
        X sum = ex;
        do{
            while(l % 2 == 0) l >>= 1;
            if(!f(fx(sum,dat[l]))){
                while(l < n){
                    push_lazy(l);
                    l *= 2;
                    if(f(fx(sum,dat[l]))){
                        sum = fx(sum,dat[l]);
                        l++;
                    }
                }
                return l - n;
            }
            sum = fx(sum,dat[l]);
            l++;
        }while((l & -l) != l);
        return N;
    }

    //x=prod(l,r),f(x)=trueとなる最小のlを求める(f(ex)=true, 0-indexed),O(logN)
    long long min_left(function<bool(X)> f, long long r = -1){
        if(r == 0) return 0;
        if(r == -1) return N;
        r += n;
        push_lazy_deep(r-1);
        X sum = ex;
        do{
            r--;
            while(r > 1 && (r % 2)) r >>= 1;
            if(!f(fx(dat[r],sum))){
                while(r < n){
                    push_lazy(r);
                    r *= 2;
                    if(f(fx(dat[r],sum))){
                        sum = fx(dat[r],sum);
                        r--;
                    }
                }
                return r + 1 - n;
            }
            sum = fx(dat[r],sum);
        }while((r & -r) != r);
        return 0;
    }
};

template <typename X>
struct SegTree{
    using FX = function<X(X,X)>; //Xを2つ受け取りXを返す関数の型
    long long n,N;
    FX fx; 
    X ex;
    vector<X> dat;

    //要素数N,二項演算fx,単位元ex;
    SegTree(long long _n, FX _fx, X _ex){
        init(_n, _fx, _ex);
    }

    void init(long long _n, FX _fx, X _ex){
        N = _n, fx = _fx; ex = _ex;
        long long x = 1;
        while(_n > x) x *= 2;
        n = x;
        dat.assign(n*2,_ex);
    }

    //i番目の要素にアクセス(0-indexed),O(1)
    X operator[](long long i){return dat[n+i];}
    X get(long long i){return dat[n+i];}

    //i番目の要素をxにアップデート(0-indexed),O(logN)
    void set(long long i, X x){
        i += n;
        dat[i] = x;
        while(i >>= 1){
            dat[i] = fx(dat[2*i],dat[2*i+1]);
        }
    }

    //[l,r)で二項演算を作用した結果(0-indexed),O(logN)
    X prod(long long l, long long r){
        X vleft = ex, vright = ex;
        for(long long left = l+n, right = r+n; left < right; left >>= 1, right >>= 1){
            if(left & 1) vleft = fx(vleft,dat[left++]);
            if(right & 1) vright = fx(dat[--right],vright);
        }
        return fx(vleft,vright);
    }
    //[0,N)まで二項演算を作用,O(1)
    X all_prod() {return dat[1];}

    //x=prod(l,r),f(x)=trueとなる最大のrを求める(f(ex)=true, 0-indexed),O(logN)
    long long max_right(const function<bool(X)> f, long long l = 0){
        if(l == N) return N;
        l += n;
        X sum = ex;
        do{
            while(l % 2 == 0) l >>= 1;
            if(!f(fx(sum,dat[l]))){
                while(l < n){
                    l = l * 2;
                    if(f(fx(sum,dat[l]))){
                        sum = fx(sum,dat[l]);
                        l++;
                    }
                }
                return l - n;
            }
            sum = fx(sum,dat[l]);
            l++;
        }while((l & -l) != l);
        return N;
    }
  
    //x=prod(l,r),f(x)=trueとなる最小のlを求める(f(ex)=true, 0-indexed),O(logN)
    long long min_left(const function<bool(X)> f, long long r = -1){
        if(r == 0) return 0;
        if(r == -1) return N;
        r += n;
        X sum = ex;
        do{
            r--;
            while(r > 1 && (r % 2)) r >>= 1;
            if(!f(fx(dat[r],sum))){
                while(r < n){
                    r = r * 2 + 1;
                    if(f(fx(dat[r],sum))){
                        sum = fx(dat[r],sum);
                        r--;
                    }
                }
                return r + 1 - n;
            }
            sum = fx(dat[r],sum);
        }while((r & -r) != r);
        return 0;
    } 
};


int main(){
    cin.tie(0);
    ios_base::sync_with_stdio(0);
    ll N,M; cin >> N >> M;
    VL a(N),l(N),r(N);
    for(ll i = 0; i < N; i++){
        cin >> a[i] >> l[i] >> r[i];
        l[i]--,r[i]--;
    }

    auto f = [](ll x1, ll x2){
        return x1+x2;
    };
    ll e = 0;
    SegTree<ll> seg(M,f,e);

    using X = PLL;
    using T = ll; 
    auto fx = [](X x1, X x2){
        return X{x1.first+x2.first,x1.second+x2.second};
    };
    X ex = {0,0};
    auto mapping = [](T t, X x){
        return X{x.first+t*x.second,x.second};
    };
    auto comp = [](T t1, T t2){
        return t1+t2;
    };
    T et = 0;
    LazySegTree<X,T> minus(M,fx,mapping,comp,ex,et),other(M,fx,mapping,comp,ex,et);

    for(ll i = 0; i < M; i++) minus.set(i,{0,1});
    for(ll i = 0; i < M; i++) other.set(i,{0,1});
    for(ll i = 0; i < N; i++) seg.set(i,a[i]);

    ll ans = 0;
    for(ll i = 0; i < N; i++){
        minus.apply(l[i],r[i]+1,1);
        other.apply(l[i],r[i]+1,a[i]);
    }

    //for(ll i = 0; i < M; i++) cout << minus[i].first << " ";
    //cout << '\n';

    for(ll i = 0; i < N; i++){
        ans += (r[i]-l[i]+1)*a[i];
        ans -= a[i]*minus[i].first;
    }
    //cout << ans << '\n';

    VL now(N,0);
    iota(ALL(now),0);

    ll Q; cin >> Q;
    while(Q--){
        ll X,Y,U,V; cin >> X >> Y >> U >> V;
        X--,Y--,U--,V--;

        ans -= (r[X]-l[X]+1)*a[X];
        ans += a[X]*minus[now[X]].first;

        //minus.apply(l[X],r[X]+1,-1);
        other.apply(l[X],r[X]+1,-a[X]);
        seg.set(now[X],0);
        {
            PLL m = other[now[X]];
            ll p = seg.prod(l[X],r[X]+1);
            //if(l[X] <= now[X] and now[X] <= r[X]) m.first -= a[X];
            ans += p;
        }

        minus.apply(l[X],r[X]+1,-1);
        minus.apply(U,V+1,1);

        ans += (V-U+1)*a[X];
        ans -= a[X]*minus[Y].first;

        {
            PLL m = other[Y];
            ll p = seg.prod(U,V+1);
            //if(l[X] <= Y and Y <= r[X]) m.first -= a[X];
            ans -= p;
        }
        //minus.apply(U,V+1,1);

        seg.set(Y,a[X]);
        other.apply(U,V+1,a[X]);

        now[X] = Y;
        l[X] = U, r[X] = V;

        cout << ans << '\n';
    }
    
    
}
0