IOI '14 P3 - Game

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Points: 20 (partial)
Time limit: 2.5s
Memory limit: 256M

Problem type
Allowed languages
C, C++

Jian-Jia is a young boy who loves playing games. When he is asked a question, he prefers playing games rather than answering directly. Jian-Jia met his friend Mei-Yu and told her about the flight network in Taiwan. There are n cities in Taiwan (numbered 0, \dots, n - 1), some of which are connected by flights. Each flight connects two cities and can be taken in both directions.

Mei-Yu asked Jian-Jia whether it is possible to go between any two cities by plane (either directly or indirectly). Jian-Jia did not want to reveal the answer, but instead suggested to play a game. Mei-Yu can ask him questions of the form "Are cities x and y directly connected with a flight?", and Jian-Jia will answer such questions immediately. Mei-Yu will ask about every pair of cities exactly once, giving r = \frac{n(n-1)} 2 questions in total. Mei-Yu wins the game if, after obtaining the answers to the first i questions for some i < r, she can infer whether or not it is possible to travel between every pair of cities by flights (either directly or indirectly). Otherwise, that is, if she needs all r questions, then the winner is Jian-Jia.

In order for the game to be more fun for Jian-Jia, the friends agreed that he may forget about the real Taiwanese flight network, and invent the network as the game progresses, choosing his answers based on Mei-Yu's previous questions. Your task is to help Jian-Jia win the game, by deciding how he should answer the questions.


We explain the game rules with three examples. Each example has n = 4 cities and r = 6 rounds of question and answer.

In the first example (the following table), Jian-Jia loses because after round 4, Mei-Yu knows for certain that one can travel between any two cities by flights, no matter how Jian-Jia answers questions 5 or 6.

round question answer
1 0, 1 yes
2 3, 0 yes
3 1, 2 no
4 0, 2 yes
----- -------- ------
5 3, 1 no
6 2, 3 no

In the next example Mei-Yu can prove after round 3 that no matter how Jian-Jia answers questions 4, 5, or 6, one cannot travel between cities 0 and 1 by flights, so Jian-Jia loses again.

round question answer
1 0, 3 no
2 2, 0 no
3 0, 1 no
----- -------- ------
4 1, 2 yes
5 1, 3 yes
6 2, 3 yes

In the final example Mei-Yu cannot determine whether one can travel between any two cities by flights until all six questions are answered, so Jian-Jia wins the game. Specifically, because Jian-Jia answered yes to the last question (in the following table), then it is possible to travel between any pair of cities. However, if Jian-Jia had answered no to the last question instead then it would be impossible.

round question answer
1 0, 3 no
2 1, 0 yes
3 0, 2 no
4 3, 1 yes
5 1, 2 no
6 2, 3 yes


Please write a program that helps Jian-Jia win the game. Note that neither Mei-Yu nor Jian-Jia knows the strategy of each other. Mei-Yu can ask about pairs of cities in any order, and Jian-Jia must answer them immediately without knowing the future questions. You need to implement the following two functions.

  • initialize(n) -- We will call your initialize first. The parameter n is the number of cities.
  • hasEdge(u, v) -- Then we will call hasEdge for r = \frac {n(n-1)} 2 times. These calls represent Mei-Yu's questions, in the order that she asks them. You must answer whether there is a direct flight between cities u and v. Specifically, the return value should be 1 if there is a direct flight, or 0 otherwise.


Each subtask consists of several games. You will only get points for a subtask if your program wins all of the games for Jian-Jia.

subtask points n
1 15 n = 4
2 27 4 \le n \le 80
3 58 4 \le n \le 1\,500

Implementation details

Your submission implements the subprograms described above using the following signatures.

void initialize(int n);
int hasEdge(int u, int v);


  • -2
    idkanything  commented on Sept. 7, 2020, 9:33 p.m.


  • 1
    Kirito  commented on Aug. 22, 2018, 12:55 p.m. edited

    zubec has pointed out the grader is incorrect, and all submissions have been rejudged.

    Credit to george_chen for identifying the bug.

  • 1
    sunnylancoder  commented on Oct. 19, 2017, 7:46 p.m.

    I don't think the math is showing up properly. I'm getting

    (4 \le n \le 1\,500)

    in the constraints

    • 1
      Kirito  commented on Oct. 20, 2017, 11:57 a.m.


    • 0
      Jeffmagma  commented on Oct. 19, 2017, 10:28 p.m.

      (4 \le n \le 80)

      (4 \le n \le 1\,500)

  • 0
    pyrexshorts  commented on July 8, 2015, 10:40 p.m.

    It should be r = (n)(n-1) / 2.