IOI 2018 – Mechanical Doll Problem and Solution

IOI 2018 – Mechanical Doll Problem and Solution

The 30th International Olympiad in Informatics was held in Tsukuba, Japan in 2018. There were two competition days, with 3 problems given to the competitors on each day. You can see Mechanical Doll problem and solutions below.

Problem :

ioi2018problem4

Solution :

#include "doll.h"
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 400005;

int vtx_number, total_leaf;
int X[MAXN], Y[MAXN], mp[MAXN];

void dfs(int s, int e, int v, int d){
	vtx_number++;
	if(e - s == 1){
		if(s < total_leaf) X[vtx_number - 1] = -1;
		else X[vtx_number - 1] = v;
		Y[vtx_number - 1] = v ^ (1 << d);
		return;
	}
	int cur_vtx = vtx_number - 1;
	int m = (s + e) / 2;
	if(m < total_leaf){
		X[cur_vtx] = -1;
	}
	else{
		X[cur_vtx] = -vtx_number - 1;
		dfs(s, m, v, d + 1);
	}
	Y[cur_vtx] = -vtx_number - 1;
	dfs(m+1, e, v ^ (1 << d), d + 1);
}

void create_circuit(int M, std::vector<int> A) {
	if(A.size() == 1){
		vector<int> C(M + 1, 0), X, Y;
		C[0] = A[0];
		answer(C, X, Y);
		return;
	}
	vector<int> C(M + 1, -1);
	C[0] = A[0];
	int K = 0;
	while((1 << K) < A.size()) K++;
	total_leaf = (1 << K) - A.size();
	A.erase(A.begin());
	A.push_back(0);
	dfs(0, (1 << K) - 1, 0, 0);
	for(int i=0; i<vtx_number; i++){
		if(X[i] >= 0) mp[X[i]] = 1;
		if(Y[i] >= 0) mp[Y[i]] = 1;
	}
	int ptr = 0;
	for(int i=0; i<(1<<K); i++){
		if(mp[i]){
			mp[i] = A[ptr++];
		}
	}
	for(int i=0; i<vtx_number; i++){
		if(X[i] >= 0) X[i] = mp[X[i]];
		if(Y[i] >= 0) Y[i] = mp[Y[i]];
	}
	vector<int> vx(X, X + vtx_number);
	vector<int> vy(Y, Y + vtx_number);
	answer(C, vx, vy);
}

Explanation :

doll-review

 55 total views,  2 views today

Post Disclaimer

The hole problem statement are given by ioinformatics.org but the solution are generated by the sltechacademy authority if any of the query regarding this post or website fill the following contact form thank you. 

Leave a Reply

Your email address will not be published. Required fields are marked *