POJ1013 Counterfeit Dollar

题目来源：http://poj.org/problem?id=1013

题目大意：有12枚硬币，其中有一枚假币。所有钱币的外表都一样，所有真币的重量都一样，假币的重量与真币不同，但我们不知道假币的重量是比真币轻还是重。现在有一个很准确的天平，我们可以用这个天平称3次来找到那枚假币。只要仔细选择三次称的方式，总可以再三次之内找出那枚假币。

输入：第一行一个正整数n表示样例个数。接下来每三行为一个测试样例。每行为一次称的结果。每枚硬币被编号为A--L。称量的结果有三种，分别用“up”、“down”和“even”表示。第一个字符串表示天平左边的硬币，第二个字符串表示右边的硬币。左边和右边的硬笔数总是相等的。第三个字符串的单词表明天平右边的状态。

输出：对于每个测试用例，输出假币的编号和这枚假币是比真币重还是轻。格式依照Sample output.

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Sample Input

1
ABCD EFGH even
ABCI EFJK up
ABIJ EFGH even 

Sample Output

K is the counterfeit coin and it is light.

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　　注意到以下几点：

　　1.某次称量天平平衡，说明天平两端都是真币.

　　2.某次天气不平衡，说明这次称量没有用到的都是真币.

　　3.如果假币比真币重，则假币只可能每次都出现在天平重的一端（轻则相反），所以若某硬币一次出现在重的一端另一次出现在轻了一端则为真币

　　给每枚硬币一个编码，表示其状态。-1表示没有出现，1表示是真币，0表示可能是假币，且比真币轻，2表示可能是假币，且比真币重。

　　依据上述观察，每称一次更新一次硬币状态。最终只有一枚硬币不为1。具体见代码。

 //////////////////////////////////////////////////////////////////////////
 //        POJ1013 Counterfeit Dollar
 //        Memory: 268K        Time: 16MS
 //        Language: C++        Result: Accepted
 //////////////////////////////////////////////////////////////////////////

 #include <iostream>
 #include <string>
 using namespace std;

 int main() {
     int n;
     cin >> n;
     for (int i = ; i < n; i++) {
         string l[], r[], b[];
         int result[];
         bool light = false;
         for (int i = ; i < ; i++) {
             result[i] = -;//未出现
         }
         cin >> l[] >> r[] >> b[] >> l[] >> r[] >> b[] >> l[] >> r[] >> b[];

         for(int i = ; i < ; i++) {
             if (b[i].compare("even") == ) {
                 for (int j = ; j < l[i].size(); j++) {
                     result[l[i][j] - 'A'] = ;
                 }
                 for (int j = ; j < r[i].size(); j++) {
                     result[r[i][j] - 'A'] = ;
                 }
             } else if (b[i].compare("up") == ) {
                 bool mark[] = {false, false, false, false, false, false,
                                  false, false, false, false, false, false};
                 for (int j = ; j < l[i].size(); j++) {
                     if (result[l[i][j] - 'A'] == -) {
                         result[l[i][j] - 'A'] = ;
                     } else if (result[l[i][j] - 'A'] == ) {
                         result[l[i][j] - 'A'] = ;
                     }
                     mark[l[i][j] - 'A'] = true;
                 }
                 for (int j = ; j < r[i].size(); j++) {
                     if (result[r[i][j] - 'A'] == -) {
                         result[r[i][j] - 'A'] = ;
                     } else if (result[r[i][j] - 'A'] == ) {
                         result[r[i][j] - 'A'] = ;
                     }
                     mark[r[i][j] - 'A'] = true;
                 }
                 for (int t =  ; t < ; t++) {
                     if (mark[t] == false) {
                         result[t] = ;
                     }
                 }
             } else {
                 bool mark[] = {false, false, false, false, false, false,
                                  false, false, false, false, false, false};
                 for (int j = ; j < l[i].size(); j++) {
                     if (result[l[i][j] - 'A'] == -) {
                         result[l[i][j] - 'A'] = ;
                     } else if (result[l[i][j] - 'A'] == ) {
                         result[l[i][j] - 'A'] = ;
                     }
                     mark[l[i][j] - 'A'] = true;
                 }
                 for (int j = ; j < r[i].size(); j++) {
                     if (result[r[i][j] - 'A'] == -) {
                         result[r[i][j] - 'A'] = ;
                     } else if (result[r[i][j] - 'A'] == ) {
                         result[r[i][j] - 'A'] = ;
                     }
                     mark[r[i][j] - 'A'] = true;
                 }
                 for (int t =  ; t < ; t++) {
                     if (mark[t] == false) {
                         result[t] = ;
                     }
                 }
             }
         }
         for (int i = ; i < ; i++) {
             if (result[i] == ) {
                 cout << char(i + 'A') << " is the counterfeit coin and it is light." << endl;
                 break;
             } else if (result[i] == ) {
                 cout << char(i + 'A') << " is the counterfeit coin and it is heavy." << endl;
                 break;
             }
         }
     }
     system("pause");
 }

附Discuss里面的一些测试数据：

 sample input 

 ABCD EFGH even
 ABCI EFJK up
 ABIJ EFGH even
 AGHL BDEC even
 JKI ADE up
 J K even
 ABCDEF GHIJKL up
 ABC DEF even
 I J down
 ABCDEF GHIJKL up
 ABHLEF GDIJKC down
 CD HA even
 A B up
 B A down
 A C even
 A B up
 B C even
 DEFG HIJL even
 ABC DEJ down
 ACH IEF down
 AHK IDJ down
 ABCD EFGH even
 AB IJ even
 A L down
 EFA BGH down
 EFC GHD even
 BA EF down
 A B up
 A C up
 L K even
 ACEGIK BDFHJL up
 ACEGIL BDFHJK down
 ACEGLK BDFHJI down
 ACEGIK BDFHJL up
 ACEGIL BDFHJK down
 ACEGLK BDFHJI up 

 sample output
 K is the counterfeit coin and it is light.
 I is the counterfeit coin and it is heavy.
 I is the counterfeit coin and it is light.
 L is the counterfeit coin and it is light.
 B is the counterfeit coin and it is light.
 A is the counterfeit coin and it is heavy.
 A is the counterfeit coin and it is light.
 L is the counterfeit coin and it is heavy.
 A is the counterfeit coin and it is light.
 A is the counterfeit coin and it is heavy.
 L is the counterfeit coin and it is light.
 K is the counterfeit coin and it is heavy.

Test data & answer