Submission #17047

#TimeUsernameProblemLanguageResultExecution timeMemory
17047erdemkirazDancing Elephants (IOI11_elephants)C++98
26 / 100
9000 ms38368 KiB
#include "elephants.h" #include <bits/stdc++.h> using namespace std; #define type(x) __typeof((x).begin()) #define foreach(it, x) for(type(x) it = (x).begin(); it != (x).end(); it++) typedef long long ll; typedef pair < int, int > ii; const int inf = 1e9 + 333; const ll linf = 1e18 + inf; const int N = 150000 + 5; const int K = 2; const int T = N / K + 5; int n, l; int a[N], gr[N], nxt[N], need[N]; set < ii > s[T], all; inline int get(int x) { return all.lower_bound(ii(x + l + 1, 0)) -> second; } void rebuild(int i) { int last = 0; if(s[i].size()) last = (--s[i].end()) -> first; for(set < ii > :: reverse_iterator it = s[i].rbegin(); it != s[i].rend(); it++) { int x = it -> first; int id = it -> second; gr[id] = i; int next = get(x); //printf("id = %d next = %d\n", id, next); //printf("id = %d next = %d\n", id, next); if(a[next] <= last) { nxt[id] = nxt[next]; need[id] = need[next] + 1; } else { nxt[id] = -1; need[id] = 0; } } } ii b[N]; void init() { all.clear(); for(int i = 0; i < n; i++) { all.insert(ii(a[i], i)); b[i] = ii(a[i], i); } a[n] = 2 * inf; all.insert(ii(2 * inf, n)); sort(b, b + n); int ind = 0; for(int i = 0; i * K < n; i++) { s[i].clear(); for(int j = i * K; j < n and j < (i + 1) * K; j++) { s[i].insert(b[j]); } ind = i; } for(int i = ind; i >= 0; i--) rebuild(i); } void init(int N, int L, int X[]) { n = N; l = L; for(int i = 0; i < n; i++) { a[i] = X[i]; } init(); } int update(int i, int x) { int ind = -1; for(int j = 0; j * K < n; j++) { if(s[j].count(ii(a[i], i))) { ind = j; break; } } assert(ind != -1); s[ind].erase(ii(a[i], i)); all.erase(ii(a[i], i)); rebuild(ind); a[i] = x; for(int j = 0; j * K < n; j++) { ind = j; if(s[j].size() and x <= (--s[j].end()) -> first) break; } all.insert(ii(a[i], i)); s[ind].insert(ii(a[i], i)); rebuild(ind); int pos = all.begin() -> second, ans = 0; while(pos < n) { //printf("pos = %d\n", pos); //getchar();getchar(); if(nxt[pos] == -1) { ans++; pos = get(a[pos]); } else { ans += need[pos]; pos = nxt[pos]; } } return ans; }
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