제출 #56090

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
56090		Benq	전선 연결 (IOI17_wiring)	C++14	100 / 100	193 ms	72620 KiB

wiring

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
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;

ll dp[200001];
set<int> S[2];

int getDist(pi x) {
    auto it = S[x.s^1].lb(x.f);
    int ret = MOD;
    if (it != S[x.s^1].end()) ret = min(ret,*it-x.f);
    if (it != S[x.s^1].begin()) ret = min(ret,x.f-*prev(it));
    return ret;
}

ll min_total_length(vi a, vi b) {
    for (int i: a) S[0].insert(i);
    for (int i: b) S[1].insert(i);
    
    vpi v;
    F0R(i,sz(a)) v.pb({a[i],0});
    F0R(i,sz(b)) v.pb({b[i],1});
    sort(all(v));
    
    ll csum = 0;
    int ind = -1;
    F0R(i,sz(v)) {
        dp[i+1] = dp[i]+getDist(v[i]);
        if (i && v[i].s != v[i-1].s) {
            csum = v[i].f-v[i-1].f;
            ind = i-1;
        } else if (ind > 0 && v[ind].s == v[ind-1].s) {
            ind --;
            csum += v[i].f-v[ind].f;
        } else ind = -1;
        
        if (ind != -1) dp[i+1] = min(dp[i+1],dp[ind]+csum);
        // cout << "HI " << i << " " << v[i].f << " " << v[i].s << " " << dp[i+1] << "\n";
    }
    
    return dp[sz(v)];
}

• Subtask 1: 7 / 7
• Subtask 2: 13 / 13
• Subtask 3: 10 / 10
• Subtask 4: 25 / 25
• Subtask 5: 45 / 45