枚举：称硬币问题

POJ1013 Counterfeit Dollar 称硬币

http://poj.org/problem?id=1013

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 47849 Accepted: 15128

---

Description

Sally Jones has a dozen Voyageur silver dollars. However, only eleven of the coins are true silver dollars; one coin is counterfeit even though its color and size make it indistinguishable from the real silver dollars. The counterfeit coin has a different weight from the other coins but Sally does not know if it is heavier or lighter than the real coins.
Happily, Sally has a friend who loans her a very accurate balance scale. The friend will permit Sally three weighings to find the counterfeit coin. For instance, if Sally weighs two coins against each other and the scales balance then she knows these two coins are true. Now if Sally weighs
one of the true coins against a third coin and the scales do not balance then Sally knows the third coin is counterfeit and she can tell whether it is light or heavy depending on whether the balance on which it is placed goes up or down, respectively.
By choosing her weighings carefully, Sally is able to ensure that she will find the counterfeit coin with exactly three weighings.

有12枚硬币。其中有11枚真币和1枚假币。假币和真币重量不同，但不知道假币比真币轻还是重。现在，用一架天平称了这些币三次，告诉你称的结果，请找出假币并且确定假币是轻是重（数据保证一能找出来）。

Input

The first line of input is an integer n (n > 0) specifying the number of cases to follow. Each case consists of three lines of input, one for each weighing. Sally has identified each of the coins with the letters A–L. Information on a weighing will be given by two strings of letters and then one of the words “up”, “down”, or “even”. The first string of letters will represent the coins on the left balance; the second string, the coins on the right balance. (Sally will always place the same number of coins on the right balance as on the left balance.) The word in the third position will tell whether the right side of the balance goes up, down, or remains even.

第一行是测试数据组数。
每组数据有三行，每行表示一次称量的结果。银币标号为A-L。每次称量的结果用三个以空格隔开的字符串表示：天平左边放置的硬币 天平右边放置的硬币 平衡状态。其中平衡状态用 “up”, “down”, 或 “even” 表示 ,分别表示右端高、右端低和平衡。天平左右的硬币数总是相等的。

Output

For each case, the output will identify the counterfeit coin by its letter and tell whether it is heavy or light. The solution will always be uniquely determined.

输出哪一个标号的银币是假币，并说明它比真币轻还是重。

Sample Input

1
ABCD EFGH even
ABCI EFJK up
ABIJ EFGH even

Sample Output

K is the counterfeit coin and it is light.

Source

East Central North America 1998

解题思路

由于只有一枚假币，我们先假设12枚硬币其中的一枚是轻的，看这样是否符合称量结果。如果符合，问题即解决。如果不符合，就假设它是重的，看是否符合称量结果，还不符合就枚举下一枚硬币。把所有硬币都试一遍，数据保证一定能找到特殊硬币。

代码实现

#include <iostream>#include <cstring>using namespace std;char Left[3][7];            //天平左边硬币，称三次，每次每边最多六枚硬币 char Right[3][7];           //天平右边硬币 char result[3][7];          //结果bool IsFake(char c, bool light);        int main(){    int t;    cin >> t;    while(t--){        for(int i = 0; i<3; ++i)            //输入三组数据             cin >> Left[i] >> Right[i] >> result[i];            //数据格式：左侧硬币 右侧硬币 右侧天平状态         for(char c='A'; c<='L';c++){        //从A开始试             if(IsFake(c,true)){             //先假设硬币是假的，转到IsFake函数进行枚举，如果返回true表明找到了假币                cout << c << " is the counterfeit coin and it is light." << endl;                break;          //题目表明一定会出现假币            }else if(IsFake(c,false)){                cout << c << " is the counterfeit coin and it is heavy." << endl;                break;            }        }       }       return 0;} bool IsFake(char c, bool light){        //light 为真表示假币为轻，为假表示假币为重      for(int i=0;i<3;++i){               //将假设带入到三次称量结果中，如果不矛盾说明假设成立        char * pLeft, *pRight;          //指向天平两边的字符串             if(light){            pLeft = Left[i];            pRight = Right[i];        }else{                          //如果假设假币是重的则要把称量结果左右对调            pLeft = Right[i];            pRight = Left[i];        }        switch(result[i][0]){                           //天平右边的情况            case 'u':                                                   if(strchr(pRight,c) == NULL)                    return false;                break;            case 'e':                if(strchr(pLeft,c)||strchr(pRight,c))                    return false;                break;            case 'd':                if(strchr(pLeft,c) == NULL)                    return false;                break;        }    }    return true;}