首页 > ACM题库 > HDU-杭电 > Hdu 1609 The Game of Master-Mind 待解决 [解题报告] C++

Hdu 1609 The Game of Master-Mind 待解决 [解题报告] C++

The Game of Master-Mind

问题描述 :

If you want to buy a new cellular phone, there are many various types to choose from. To decide which one is the best for you, you have to consider several important things: its size and weight, battery capacity, WAP support, colour, price. One of the most important things is also the list of games the phone provides. Nokia is one of the most successful phone makers because of its famous Snake and Snake II. ACM wants to make and sell its own phone and they need to program several games for it. One of them is Master-Mind, the famous board logical game.

The game is played between two players. One of them chooses a secret code consisting of P ordered pins, each of them having one of the predefined set of C colours. The goal of the second player is to guess that secret sequence of colours. Some colours may not appear in the code, some colours may appear more than once.

The player makes guesses, which are formed in the same way as the secret code. After each guess, he/she is provided with an information on how successful the guess was. This feedback is called a hint. Each hint consists of B black points and W white points. The black point stands for every pin that was guessed right, i.e. the right colour was put on the right position. The white point means right colour but on the wrong position. For example, if the secret code is "white, yellow, red, blue, white" and the guess was "white, red, white, white, blue", the hint would consist of one black point (for the white on the first position) and three white points (for the other white, red and blue colours). The goal is to guess the sequence with the minimal number of hints.

The new ACM phone should have the possibility to play both roles. It can make the secret code and give hints, but it can also make its own guesses. Your goal is to write a program for the latter case, that means a program that makes Master-Mind guesses.


There is a single positive integer T on the first line of input. It stands for the number of test cases to follow. Each test case describes one game situation and you are to make a guess. On the first line of each test case, there are three integer numbers, P, C and M. P ( 1 <= P <= 10) is the number of pins, C (1 <= C <= 100) is the number of colours, and M (1 <= M <= 100) is the number of already played guesses.

Then there are 2 x M lines, two lines for every guess. At the first line of each guess, there are P integer numbers representing colours of the guess. Each colour is represented by a number Gi, 1 <= Gi <= C. The second line contains two integer numbers, B and W, stating for the number of black and white points given by the corresponding hint.

Let’s have a secret code S1, S2, … SP and the guess G1, G2, … GP. Then we can make a set H containing pairs of numbers (I,J) such that SI = GJ, and that any number can appear at most once on the first position and at most once on the second position. That means for every two different pairs from that set, (I1,J1) and (I2,J2), we have I1 <> I2 and J1 <> J2. Then we denote B(H) the number of pairs in the set, that meet the condition I = J, and W(H) the number of pairs with I <> J.

We define an ordering of every two possible sets H1 and H2. Let’s say H1 <= H2 if and only if one of the following holds:

B(H1) < B(H2), or
B(H1) = B(H2) and W(H1) <= W(H2)

Then we can find a maximal set Hmax according to this ordering. The numbers B(Hmax) and W(Hmax) are the black and white points for that hint.


For every test case, print the line containing P numbers representing P colours of the next guess. Your guess must be valid according to all previous guesses and hints. The guess is valid if the sequence could be a secret code, i.e. the sequence was not eliminated by previous guesses and hints.

If there is no valid guess possible, output the sentence You are cheating!. If there are more valid guesses, output the one that is lexicographically smallest. I.e. find such guess G that for every other valid guess V there exists such a number I that:

GJ = VJ for every J<I, and


4 3 2
1 2 3 2
1 1
2 1 3 2
1 1
4 6 2
3 3 3 3
3 0
4 4 4 4
2 0
8 9 3
1 2 3 4 5 6 7 8
0 0
2 3 4 5 6 7 8 9
1 0
3 4 5 6 7 8 9 9
2 0


1 1 1 3
You are cheating!
9 9 9 9 9 9 9 9

  1. 思路二可以用一个长度为k的队列来实现,入队后判断下队尾元素的next指针是否为空,若为空,则出队指针即为所求。

  2. 一开始就规定不相邻节点颜色相同,可能得不到最优解。我想个类似的算法,也不确定是否总能得到最优解:先着一个点,随机挑一个相邻点,着第二色,继续随机选一个点,但必须至少有一个边和已着点相邻,着上不同色,当然尽量不增加新色,直到完成。我还找不到反例验证他的错误。。希望LZ也帮想想, 有想法欢迎来邮件。谢谢