UVa 340 Master-Mind Hints

蛋疼的题目描述，看了好长好长时间才看懂，题目本身是很简单的。

Designer给出一串长度为N的Code，Breaker用Guess来破译。

对于两串数字，如果有同一列相等的数字，那么叫做strong  match，

位于不同列的相等的两个数字，叫做weak  match。

题目要求就是先输出strong的个数，然后是weak的个数。

对了，需要注意的是

1、每个code只能匹配一次，不论是strong还是match。

2、输出（*，*）的时候注意前面那四个空格，就因为这个我还PE了一次。

如果还不明白，请看这里

再说说我的算法：我又开了两个数组num1，num2来分别放code和guess中1~9每个数字出现的次数。首先匹配strong，匹配好的同时将num1和num2中对应的元素减1，表示已经匹配过了。strong匹配完以后，剩下的就是weak。因为已经不存在strong的情况，所以weak直接+=min（code， guess）。

以下是原题：

 Master-Mind Hints 

MasterMind is a game for two players. One of them, Designer, selects a secret code. The other, Breaker, tries to break it. A code is no more than a row of colored dots. At the beginning of a game, the players agree upon the length N that a code must have and upon the colors that may occur in a code.

In order to break the code, Breaker makes a number of guesses, each guess itself being a code. After each guess Designer gives a hint, stating to what extent the guess matches his secret code.

In this problem you will be given a secret code  and a guess  , and are to determine the hint. A hint consists of a pair of numbers determined as follows.

A match is a pair (i,j),  and  , such that  . Match (i,j) is called strong when i =j, and is called weak otherwise. Two matches (i,j) and (p,q) are called independent when i = p if and only if j = q. A set of matches is called independent when all of its members are pairwise independent.

Designer chooses an independent set M of matches for which the total number of matches and the number of strong matches are both maximal. The hint then consists of the number of strong followed by the number of weak matches in M. Note that these numbers are uniquely determined by the secret code and the guess. If the hint turns out to be (n,0), then the guess is identical to the secret code.

Input

The input will consist of data for a number of games. The input for each game begins with an integer specifying N (the length of the code). Following these will be the secret code, represented as N integers, which we will limit to the range 1 to 9. There will then follow an arbitrary number of guesses, each also represented as N integers, each in the range 1 to 9. Following the last guess in each game will be N zeroes; these zeroes are not to be considered as a guess.

Following the data for the first game will appear data for the second game (if any) beginning with a new value for N. The last game in the input will be followed by a single zero (when a value for N would normally be specified). The maximum value for N will be 1000.

Output

The output for each game should list the hints that would be generated for each guess, in order, one hint per line. Each hint should be represented as a pair of integers enclosed in parentheses and separated by a comma. The entire list of hints for each game should be prefixed by a heading indicating the game number; games are numbered sequentially starting with 1. Look at the samples below for the exact format.

Sample Input

4
1 3 5 5
1 1 2 3
4 3 3 5
6 5 5 1
6 1 3 5
1 3 5 5
0 0 0 0
10
1 2 2 2 4 5 6 6 6 9
1 2 3 4 5 6 7 8 9 1
1 1 2 2 3 3 4 4 5 5
1 2 1 3 1 5 1 6 1 9
1 2 2 5 5 5 6 6 6 7
0 0 0 0 0 0 0 0 0 0
0

Sample Output

Game 1:
    (1,1)
    (2,0)
    (1,2)
    (1,2)
    (4,0)
Game 2:
    (2,4)
    (3,2)
    (5,0)
    (7,0)

AC代码：

 //#define LOCAL
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;

 const int maxn =  + ;
 int code[maxn], guess[maxn];
 int num1[], num2[], num3[];//存放code和guess中每个数字出现的次数
                                  //num3来放num1的拷贝

 int main(void)
 {
     #ifdef LOCAL
         freopen("340in.txt", "r", stdin);
     #endif

     int N, kase = ;
     while(scanf("%d", &N) ==  && N)
     {
         memset(code, , sizeof(code));
         memset(num1, , sizeof(num1));
         int i;
         for(i = ; i < N; ++i)
         {
             scanf("%d", &code[i]);
             ++num1[ code[i] ];
         }
         printf("Game %d:\n", ++kase);
         while(true)
         {
             memset(guess, , sizeof(guess));
             memset(num2, , sizeof(num2));
             memcpy(num3, num1, sizeof(num1));
             for(i = ; i < N; ++i)
             {
                 scanf("%d", &guess[i]);
                 ++num2[ guess[i] ];
             }
             if(guess[] == )
                 break;
             //统计strong的个数
             int strong = ;
             for(i = ; i < N; ++i)
             {
                 if(code[i] == guess[i])
                 {
                     ++strong;
                     --num3[ code[i] ];//每个code[i]只能匹配一次
                     --num2[ guess[i] ];
                 }
             }
             //计算weak
             int weak = ;
             for(i = ; i <= ; ++i)
                 weak += min(num3[i], num2[i]);
             printf("    (%d,%d)\n", strong, weak);
         }
     }
     return ;
 }

代码君