| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 275880 | khangal | Arranging Shoes (IOI19_shoes) | C++14 | 0 ms | 0 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
typedef long long ll;
typedef double db;
typedef pair<ll,ll> pl;
typedef vector<ll> vl;
typedef vector<vl> vvl;
#define po pop_back
#define pb push_back
#define mk make_pair
#define mt make_tuple
#define lw lower_bound
#define up upper_bound
#define ff first
#define ss second
#define BOOST ios_base::sync_with_stdio(); cin.tie(0); cout.tie(0);
#define MOD 1000000007
#define MAX 1e18
#define MIN -1e18
#define rep(i,a,b) for(ll i=a;i<=b;i++)
#define per(i,a,b) for(ll i=b;i>=a;i--)
#define con continue
#define freopen freopen("input.txt", "r", stdin);freopen("output.txt", "w", stdout);
#define PI 3.14159265358979323846264338327950288419716939937510582097494459230781640628
#define read(x) scanf("%lld",&x);
#define print(x) printf("%lld ",x);
#define endl '\n';
// typedef tree<ll , null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
// template< typename T>
// using indexed_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
ll n,m,ans,mid,mn,mx,cnt,T,sum,h1,h2,e[1234567],b[1234567],c[1234567],d[1<<20],k,i,j,l,r,h,a[1234567],w,x,y,z;
bool used[1234567];
vector<int> v[1234567],vec,vv1,vv2;
string s1,s;
int sz[1234567],par[1234567];
int tr[1234567];
ll dx[4]={-1,1,0,0},dy[4]={0,0,-1,1},c1[123][123];
map<int,vector<int>> mp,mp1;
void Add(ll x, ll val){
while(x <= 2 * n){
tr[a] += val;
x += (x&-x);
}
}
ll Sum(ll x){
ll res = 0;
while(x){
res +=tr[x];
x -= (x&-x);
}
return res;
}
int count_swaps(vector<int> vec){
n=vec.size()/2;
rep(i,0,vec.size()-1){
if(vec[i]<0)mp[vec[i]*-1].pb(i);
else mp1[vec[i]].pb(i);
}
vector<pair<int,int>>vpl;
for(auto u:mp){
x=u.ff;
y = mp[x].size();
for(ll i=0;i<y;i++){
if(mp[x][i]>mp1[x][i])ans++,swap(mp[x][i],mp1[x][i]);
vpl.pb({mp[x][i],mp1[x][i]});
}
}
for(auto u:vpl){
x=u.ff;
y=u.ss;
ans+=y-x;
ans-=(Sum(y)-Sum(x));
Add(y,1);
Add(x,1);
}
return ans;
}