제출 #698026

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
698026		sunwukong123	전선 연결 (IOI17_wiring)	C++14	100 / 100	149 ms	152684 KiB

wiring

#include "wiring.h"
#include <bits/stdc++.h>
using namespace std;
#define int long long 
#define FOR(i,s,e) for(int i = s; i <= (int)e; ++i)
#define DEC(i,s,e) for(int i = s; i >= (int)e; --i)
#define IAMSPEED ios_base::sync_with_stdio(false); cin.tie(0);
#ifdef LOCAL
#define db(x) cerr << #x << "=" << x << "\n"
#define db2(x, y) cerr << #x << "=" << x << " , " << #y << "=" << y << "\n"
#define db3(a,b,c) cerr<<#a<<"="<<a<<","<<#b<<"="<<b<<","<<#c<<"="<<c<<"\n"
#define dbv(v) cerr << #v << ":"; for (auto ite : v) cerr << ite << ' '; cerr <<"\n"
#define dbvp(v) cerr << #v << ":"; for (auto ite : v) cerr << "{"  << ite.f << ',' << ite.s << "} "; cerr << "\n"
#define dba(a,ss,ee) cerr << #a << ":"; FOR(ite,ss,ee) cerr << a[ite] << ' '; cerr << "\n"
#define reach cerr << "LINE: " << __LINE__ << "\n";
#else
#define reach 
#define db(x)
#define db2(x,y)
#define db3(a,b,c)
#define dbv(v)
#define dbvp(v)
#define dba(a,ss,ee)
#endif
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define pb push_back
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
#define f first 
#define s second
#define g0(x) get<0>(x)
#define g1(x) get<1>(x)
#define g2(x) get<2>(x)
#define g3(x) get<3>(x)
typedef pair <int, int> pi;
typedef tuple<int,int,int> ti3;
typedef tuple<int,int,int,int> ti4;
int rand(int a, int b) { return a + rng() % (b-a+1); }
const int MOD = 1e9 + 7;
const int inf = (int)1e9 + 500;
const long long oo = (long long)1e18 + 500;
template <typename T> bool chmax(T& a, const T b) { return a<b ? a = b, 1 : 0; }
template <typename T> bool chmin(T& a, const T b) { return a>b ? a = b, 1 : 0; }
const int MAXN = 200005;
int a,b;
deque<int> DQ[MAXN];
int M[MAXN];
int T[MAXN];
vector<int> dp[MAXN]; // j things from the ith group is still unpaired
vector<pi> vec;
int ans;
int arr[MAXN];
int ls[MAXN], rs[MAXN];
long long min_total_length(std::vector<int32_t> RR, std::vector<int32_t> BB){
	a=RR.size();
	b=BB.size();
	FOR(i,0,a-1){
		vec.pb({RR[i],0});
	}
	FOR(i,0,b-1) {
		vec.pb({BB[i],1});
	}
	sort(all(vec)); // {x position, 0/1}
	FOR(i,1,vec.size()) {
		arr[i]=vec[i-1].f;
	}
	int s=vec.size();
	int dlen=0;
	T[0]=-1;
	FOR(i,0,s-1) {
		if(T[dlen] != vec[i].s) {
			++dlen;
			T[dlen]=vec[i].s;
		}
		DQ[dlen].pb(vec[i].f); // all the positions of this segment
	}
	
	int idx = 1;
	FOR(i,0,dlen) {
		db(i);
		ls[i]=idx;
		idx+=DQ[i].size();
		rs[i]=idx-1;

		dp[i].resize(DQ[i].size()+3, oo);
	}
	
	dp[1][DQ[1].size()] = 0;
	FOR(i,1,dlen) { // considering i
		chmin(dp[i][DQ[i].size()], dp[i-1][0]);
		/*
		first transition: connect some things from the left in the (i-1) segment with the left most point in the ith segment
		dp[i-1][j]=dp[i-1][j+1] + (arr[lf[i]] - arr[rs[i-1]-j])
		*/
		db(i);
		dba(dp[i],0,DQ[i].size());
		DEC(j,DQ[i-1].size()-1,0){ 
			db2(arr[ls[i]] , arr[rs[i-1]-j]);
			chmin(dp[i-1][j], dp[i-1][j+1] + (arr[ls[i]] - arr[rs[i-1]-j]));
		}
		
		/* second transition: connect some things from the left in the (i-1) segment with some things from the left in the ith segment
		what is the formula? 
		just keep track of the prefix sum
		*/
		//pair the last l unpaired from previous group with this group. 
		db(i);
		dba(dp[i-1],0,DQ[i-1].size());
		int cost = 0;
		FOR(j,1,min((int)DQ[i-1].size(),(int)DQ[i].size())) {
			
			cost += arr[ls[i] + j - 1] - arr[rs[i-1] - j + 1];
			db2(dp[i-1][j] , cost);
			chmin(dp[i][DQ[i].size() - j], dp[i-1][j] + cost);
		}
		
		db(i);
		dba(dp[i],0,DQ[i].size());

		/* third transition: pair the last things here with right most element of previous group */
		if(i>1){
			DEC(j,DQ[i].size()-1,0) {
				chmin(dp[i][j], dp[i][j+1] + (arr[rs[i]-j] - arr[rs[i-1]]));
			}
		}
		
		db(i);
		dba(dp[i],0,DQ[i].size());

	}
	return dp[dlen][0];
}

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