結果
問題 |
No.3265 地元に帰れば天才扱い!
|
ユーザー |
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提出日時 | 2025-09-06 15:33:37 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 1,951 ms / 2,500 ms |
コード長 | 3,864 bytes |
コンパイル時間 | 4,850 ms |
コンパイル使用メモリ | 261,192 KB |
実行使用メモリ | 28,260 KB |
最終ジャッジ日時 | 2025-09-06 15:34:25 |
合計ジャッジ時間 | 47,140 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 4 |
other | AC * 21 |
ソースコード
#include <bits/stdc++.h> #include <atcoder/all> using namespace std; using namespace atcoder; using mint = modint998244353; //using mint = modint1000000007; using ll = long long; ll INF = 2e18; template<typename T> using vc = vector<T>; template<typename T> using vv = vc<vc<T>>; using vi = vc<int>; using vvi = vv<int>; using vl = vc<ll>; using vvl = vv<ll>; using vs = vc<string>; using vvs = vv<string>; using vb = vc<bool>; using vvb = vv<bool>; using vmint = vc<mint>; using vvmint = vv<mint>; #define rep(i,n) for(ll i=0; i<(n); i++) #define drep(i,n) for(ll i=(n)-1; i>=0; i--) #define rrep(i,n) for(ll i=1; i<=(n); i++) #define nfor(i,a,b) for(ll i=a;i<b;i++) #define dfor(i,a,b) for(ll i=(a)-1; i>=(b); i--) template<class T>istream& operator>>(istream& i, vc<T>& v) {rep(j,(ll) size(v))i >> v[j]; return i; } #define nall(a) a.begin(),a.end() #define rall(a) a.rbegin(),a.rend() template<class T> bool chmax(T& a, const T& b){ if(a < b){ a = b; return 1; } return 0; } template<class T> bool chmin(T& a, const T& b){ if(a > b){ a = b; return 1; } return 0; } #define YES cout<<"Yes"<<endl #define NO cout<<"No"<<endl #define YN {YES;}else{NO;} #define ERROR cout<<-1<<endl void print(long double x){ printf("%.20Lf\n",x);} #define vc_cout(v){ll n = size(v);rep(i,n)cout<<v[i]<<endl;} #define vv_cout(v){ll n = size(v);rep(i,n){rep(j,size(v[i])){cout<<' '<<v[i][j];}cout<<endl;}} static const ll INF64 = (1LL<<62); static const ll NINF64 = -(1LL<<62); // ============================================================= // 1. 区間加算 + 区間最小 // ============================================================= struct SMin { ll mn; }; using FAdd = ll; // 遅延作用は +f SMin op_minseg(SMin a, SMin b){ return {min(a.mn, b.mn)}; } SMin e_minseg(){ return { INF64 }; } SMin mapping_min(FAdd f, SMin x){ return {x.mn + f}; } FAdd composition_add(FAdd f, FAdd g){ return f + g; } FAdd id_add(){ return 0LL; } using LazyAddMin = lazy_segtree< SMin, op_minseg, e_minseg, FAdd, mapping_min, composition_add, id_add >; struct P_min_ge { ll T; bool operator()(SMin x) const { return x.mn >= T; } }; // ============================================================= // 3. 区間加算 + 区間和 // ============================================================= struct SAddSum { ll sum; int sz; }; SAddSum op_addsum(SAddSum a, SAddSum b){ return {a.sum + b.sum, a.sz + b.sz}; } SAddSum e_addsum(){ return {0LL, 0}; } SAddSum mapping_addsum(FAdd f, SAddSum x){ return {x.sum + f * x.sz, x.sz}; } using LazyAddSum = lazy_segtree< SAddSum, op_addsum, e_addsum, FAdd, mapping_addsum, composition_add, id_add >; struct P_sum_le_ll { ll K; bool operator()(SAddSum x) const { return x.sum <= K; } }; int main(){ int N, M; cin >> N >> M; vc<ll> A(N); vi L(N),R(N); rep(i,N) { cin >> A[i] >> L[i] >> R[i]; L[i]--; } vi live(N); rep(i,N) live[i] = i; vc<SMin> in_range(M,{0}); LazyAddMin segMin(in_range); rep(i,N) { segMin.apply(L[i],R[i],1); } vc<SAddSum> rate(M,{0,1}); LazyAddSum segSum(rate); rep(i,N) { segSum.set(i,{A[i],1}); } ll ans = 0; rep(i,N) { ans -= segSum.prod(L[i],R[i]).sum; ans += segSum.prod(L[i],R[i]).sz * A[i]; } int Q; cin >> Q; rep(q,Q) { int x,ny,u,v; cin >> x >> ny >> u >> v; x--,ny--,u--; int y = live[x]; live[x] = ny; ans += A[x] * ((v-u) - (R[x] - L[x])); ans += A[x] * segMin.get(y).mn; segMin.apply(L[x],R[x],-1); segMin.apply(u,v,1); ans -= A[x] * segMin.get(ny).mn; segSum.set(y,{0,1}); ans -= segSum.prod(u,v).sum - segSum.prod(L[x],R[x]).sum; segSum.set(ny,{A[x],1}); cout << ans << endl; L[x] = u; R[x] = v; } }