| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 609483 | HediChehaidar | 전선 연결 (IOI17_wiring) | C++17 | 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 "wiring.h"
#include<bits/stdc++.h>
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef double db;
ll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;} // greatest common divisor (PGCD)
ll lcm(ll a , ll b) {return (a * b) / gcd(a , b);} // least common multiple (PPCM)
#define ss second
#define ff first
#define all(x) (x).begin() , (x).end()
#define pb push_back
#define vi vector<int>
#define vii vector<pair<int,int>>
#define vl vector<ll>
#define vll vector<pair<ll,ll>>
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pdd pair<double,double>
#define vdd vector<pdd>
#define dte tuple<double , double , double>
using namespace std;
const int INF = 1000*1000*1000; // 1 e 9
const int MOD = 1e9 + 7;//998244353 ;
const double EPS = 0.000000001; // 1 e -9
const ll inf = (ll)1e18;
vl r , b;
int n , m;
vl v;
int col[(int)2e5 + 10];
ll dp[(int)2e5 + 10][(1 << 6)];
ll solve(int pos , int mask){
if(pos == n + m) return (mask == (1 << 6) - 1) ? 0 : inf;
if(pos >= 6 && !(mask & (1 << 5))){
return col[pos] == col[pos - 6] ? inf : (solve(pos + 1 , 2 * mask + 1)+ v[pos] - v[pos - 6] );
}
if(dp[pos][mask] != -1) return dp[pos][mask];
int x = pos >= 6 ? mask - (1 << 5) : mask;
ll ans = solve(pos + 1 , 2 * x);
for(int i = 0 ; i < min(pos , 6) ; i++){
if(col[pos] != col[pos - i - 1]){
x = pos >= 6 ? mask - (1 << 5) : mask;
assert(x >= 0);
if(i != 5) x = x | (1 << i);
ans = min(ans , v[pos] - v[pos - i-1] + solve(pos + 1 , 2 * x + 1));
}
}
//if(pos == 6 && mask == (1 << 6) - 1) cout << ans << "\n";
/*cout << pos << " ";
for(int i = 0 ; i < 6 ; i++) cout << ((mask & (1 << i)) != 0) << " ";
cout << " " << ans << "\n";*/
return dp[pos][mask] = ans;
}
ll min_total_length(vi R, vi B){
n = R.size(); m = B.size();
r.clear(); b.clear();
for(int i = 0 ; i < n ; i++) r.pb(R[i]);
for(int i = 0 ; i < m ; i++) b.pb(B[i]);
memset(dp , -1 , sizeof dp);
int i = 0 , j = 0;
while(i < n || j < m){
if(i == n || (j < m && b[j] < r[i])){
col[v.size()] = 1;
v.pb(b[j]); j++;
}
else{
col[v.size()] = 0; v.pb(r[i]); i++;
}
}
//for(auto c : v) cout << c << " "; cout << "\n";
return solve( 1 , 0);
}
int main(){
//ifstream fin ("testing.txt");
//ofstream fout ("output.txt");
ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int nn , mm; cin>>nn>>mm;
vi rr(nn) , bb(mm);
for(int i = 0 ; i < nn ; i++) cin>>rr[i];
for(int i = 0 ; i < mm ; i++) cin>>bb[i];
cout << min_total_length(rr , bb);
return 0;
}
/*
Think of : BS / DFS / BFS / SSSP / SCC / MSP / MAX FLOW / TOPSORT / LCA / MATRIX / DP(bitmask) / 2 POINTERS / SEG TREE / MATH / UN FIND / MO / HLD
Read the statement CAREFULLY !!
Make a GREADY APPROACH !!!! (start from highest / lowest)
Make your own TESTS !!
Be careful from CORNER CASES !
*/