Submission #1001278

#TimeUsernameProblemLanguageResultExecution timeMemory
1001278TobCultivation (JOI17_cultivation)C++14
0 / 100
1 ms348 KiB
#include <bits/stdc++.h> #pragma GCC optimize("Ofast") #define F first #define S second #define all(x) x.begin(), x.end() #define pb push_back #define FIO ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0) using namespace std; typedef long long ll; typedef pair <int, int> pii; typedef pair <ll, ll> pll; const int N = 307, inf = 2e9; int n, m, k, siz; int a[N], b[N]; int mem[2*N][N]; pii gr[2*N][N]; array <int,3> vv[2*N]; int q[6*N], dq[6*N]; vector <int> add[2*N], rem[2*N]; struct MQ { int type, l1, r1, l2, r2; inline void push(int x) {q[++r1] = x; while (l2 <= r2 && dq[r2] < x) r2--; dq[++r2] = x;} inline void pop() {if (dq[l2] == q[l1]) l2++; l1++;} inline int get() {return dq[l1];} inline void reset() {l1 = l2 = type*2*N; r1 = r2 = l1-1;} } m1, m2, m3; int main () { FIO; m1.type = 0; m2.type = 1; m3.type = 2; cin >> n >> m >> k; vector <pii> so; for (int i = 0; i < k; i++) { cin >> a[i] >> b[i]; a[i]--; b[i]--; so.pb({a[i], b[i]}); } sort(all(so)); for (int i = 0; i < k; i++) a[i] = so[i].F, b[i] = so[i].S; int d = a[0]; for (int i = 0; i < k; i++) a[i] -= d; d = n-a[k-1]-1; for (int i = 1; i < k; i++) d = max(d, a[i]-a[i-1]-1); a[k] = n-1; vector <ll> v; v.pb(d); for (int i = 0; i <= k; i++) { for (int j = i; j <= k; j++) { if (a[j]-a[i]-1 >= d) v.pb(a[j]-a[i]-1); if (a[j]+n-a[i]-1 >= d) v.pb(a[j]+n-a[i]-1); if (a[i]+n-a[j]-1 >= d) v.pb(a[i]+n-a[j]-1); } if (n-a[i]-1 >= d) v.pb(n-a[i]-1); } sort(all(v)); v.erase(unique(all(v)), v.end()); //PRECOMPUTE -> O(N^2 log N) for (int i = 0; i < k; i++) { set <int> s; multiset <int> rj; rj.insert(0); for (int j = i; j >= 0; j--) { mem[i][j] = mem[i][j+1]; gr[i][j].F = gr[i][j+1].F; gr[i][j].S = gr[i][j+1].S; if (s.find(b[j]) != s.end()) continue; auto p1 = s.lower_bound(b[j]); if (p1 == s.end()) { if (!s.empty()) rj.insert(b[j]-*(--p1)-1); s.insert(b[j]); gr[i][j] = {*s.begin(), m-*s.rbegin()-1}; } else if (p1 == s.begin()) { rj.insert(*p1-b[j]-1); s.insert(b[j]); gr[i][j] = {*s.begin(), m-*s.rbegin()-1}; } else { int x = *p1; --p1; int y = *p1; rj.erase(rj.find(x-y-1)); rj.insert(x-b[j]-1); rj.insert(b[j]-y-1); s.insert(b[j]); } mem[i][j] = *rj.rbegin(); } } //SOLVE O(N^3) ll mn = inf; for (auto x : v) { m1.reset(); m2.reset(); m3.reset(); siz = 0; int l = 0, r = -1, z = 0; for (int i = 0; i < k; i++) { int gt = (x+a[z]+1 < 2LL*n) ? x+a[z]+1 : 2*n; while (a[z]+x < a[i]) { if (vv[siz-1][2] != gt) {vv[siz++] = {l+1, r, gt}; l++;} else vv[siz-1][0]++, l++; z++; } if (!siz || vv[siz-1][2] != a[i]) {vv[siz++] = {l, r+1, a[i]}; r++;} else vv[siz-1][1]++, r++; } int gt = (x+a[z]+1 < 2LL*n) ? x+a[z]+1 : 2*n; while (z < k) { if (vv[siz-1][2] != gt) {vv[siz++] = {l+1, r, gt}; l++;} else vv[siz-1][0]++, l++; z++; } z = 0; #define C (x+max(m1.get(), m2.get()+m3.get())) for (int i = 0; i+1 < siz && vv[i][2] < 2*n; i++) { l = vv[i][0]; r = vv[i][1]; m1.push(mem[r][l]); m2.push(gr[r][l].F); m3.push(gr[r][l].S); while (vv[z][2]+n-1 < vv[i+1][2]) { mn = min(mn, C); m1.pop(); m2.pop(); m3.pop(); z++; } } } cout << mn << "\n"; return 0; }
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