제출 #372988

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
372988		Matteo_Verz	슈퍼트리 잇기 (IOI20_supertrees)	C++17	21 / 100	267 ms	24300 KiB

supertrees

#include "supertrees.h"
#include <bits/stdc++.h>

using namespace std;

int construct(vector<vector<int>> p) {
  int n = p.size();
  vector<vector<int>> answer;
  answer.resize(n);
  for(int i = 0; i < n; i++) answer[i].resize(n);

  vector <int> comp;
  // last subtask, p[i][j] == 3
  for(int i = 0; i < n; i++) {
    for(int j = 0; j < n; j++)
      if(p[i][j] > 2) return 0;
    if(p[i][i] == 2 || p[i][i] == 0) return 0;
  }

  vector <bool> isroot;
  int countcomp(0);

  comp.resize(n);
  isroot.resize(n);

  for(int i = 0; i < n; i++) { // we consider i to be the root of the connected component
    if(comp[i] != 0) continue; // it can't be a root
    comp[i] = ++countcomp;
    isroot[i] = true;

    for(int j = 0; j < n; j++) {
      if(i == j) continue;
      if(p[i][j] == 1) { // if there is only 1 edge from i to j
        comp[j] = countcomp;
        answer[i][j] = answer[j][i] = 1;
      }
    }
  }

  // now we connect the roots between them to form a cycle
  // we connect the i-th root with the i+1-th root
  for(int i = 0; i < n; i++) {
    if(isroot[i] == true) {
      for(int j = i + 1; j < n; j++) {
        if(p[i][j] == 2 && isroot[j]) {
          answer[i][j] = answer[j][i] = 1;
          j = n;
        }
      }
    }
  }

  // we connect the last with the first root
  for(int i = 0; i < n; i++) {
    if(isroot[i] == true) {
      for(int j = n - 1; j > i; j--) {
        if(p[i][j] == 2 && isroot[j]) {
          answer[i][j] = answer[j][i] = 1;
          j = 0;
        }
      }
      i = n;
    }
  }

  // now we need to verify if our build is fine
  // 1. We can't have a cycle of length 2
  if(countcomp == 2) {
    for(int i = 0; i < n; i++)
      for(int j = 0; j < n; j++)
        if(p[i][j] == 2)
          return 0;
  }


  for(int i = 0; i < n; i++) {
    for(int j = 0; j < n; j++)
      if(comp[i] != comp[j] && p[i][j] == 1)
        return 0;
      else if(comp[i] == comp[j] && (p[i][j] == 0 || p[i][j] == 2))
        return 0;
  }

  build(answer);
  return 1;
}

• Subtask 1: 11 / 11
• Subtask 2: 10 / 10
• Subtask 3: 0 / 19
• Subtask 4: 0 / 35
• Subtask 5: 0 / 21
• Subtask 6: 0 / 4