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>
#define int long long
//#define double long double
#define Nanase_Kurumi_aka_menhera_chan_is_mine ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define pb push_back
#define pi pair<double,int>
#define ALL(i) i.begin(),i.end()
#define gcd(i,j) __gcd(i,j)
#define fi first
#define se second
#define eps 0.00000001
#define ist insert
#define DNE nullptr
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC optimize("O2")
int max(int x,int y){return x>=y?x:y;}
int min(int x,int y){return x>=y?y:x;}
using namespace std;
typedef int ll;
const int N=1005;
const int M=1000005;
const int MOD=1000000007;//998244353;
const int INF=1000000000000000000;//2147483647;
int n,ans;
pi p[N];
int dp[2*N][2*N];
inline void sol(){
cin >>n;
for (int i=1;i<=2*n;i++)
cin >>p[i].fi>>p[i].se;
sort(p+1,p+2*n+1);
for (int i=1;i<=2*n;i++){
dp[0][i]=dp[0][i-1]+abs(p[i].fi-i)+abs(2-p[i].se);
dp[i][0]=dp[i-1][0]+abs(p[i].fi-i)+abs(1-p[i].se);
for (int j=1;j<i;j++)
dp[j][i-j]=min(dp[j-1][i-j]+abs(p[i].fi-j)+abs(1-p[i].se),dp[j][i-j-1]+abs(p[i].fi-(i-j))+abs(2-p[i].se));
}
cout <<dp[n][n];
}
signed main(){
Nanase_Kurumi_aka_menhera_chan_is_mine
int _=1;
//cin >>_;
while (_--) sol();
return 0;
}
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