# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
73027 | Benq | Ranklist Sorting (BOI07_sorting) | C++14 | 1070 ms | 648 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/rope>
using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;
int n;
vi s;
void compress() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> n; s.resize(n);
map<int,int> m;
F0R(i,n) {
cin >> s[i];
m[s[i]] = 0;
}
int co = 0;
for (auto& a: m) a.s = co++;
for (int& i: s) i = n-1-m[i];
}
pair<int,vpi> test(int z) {
vpi S; F0R(i,sz(s)) S.pb({s[i],1});
int lst = -1;
F0R(i,n) {
if (z&(1<<i)) S[i].s = 0;
else {
if (S[i].f < lst) return {MOD,{}};
lst = S[i].f;
}
}
vpi v;
int cur = 0;
F0Rd(i,n) {
int ind = 0; while (S[ind].f != i) ind ++;
if (S[ind].s) continue;
cur += ind+1; S.erase(S.begin()+ind);
int lst = 0;
F0R(j,n-1) if (S[j].s && S[j].f < i) lst = j+1;
cur += lst+1; S.insert(S.begin()+lst,{i,1});
v.pb({ind+1,lst+1});
}
/*for (auto a: S) cout << a.f << " " << a.s << " | ";
cout << "\n";*/
return {cur,v};
}
int main() {
compress();
pair<int,vpi> ans = {MOD,{}};
F0R(i,1<<n) ans = min(ans,test(i));
cout << sz(ans.s) << "\n";
for (auto a: ans.s) cout << a.f << " " << a.s << "\n";
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
* if you have no idea just guess the appropriate well-known algo instead of doing nothing :/
*/
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