POJ 1013 Counterfeit Dollar（枚举）

Counterfeit Dollar

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.
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.
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.
Sample Input

1
ABCD EFGH even
ABCI EFJK up
ABIJ EFGH even

Sample Output

K is the counterfeit coin and it is light.

题目大意：有12枚硬币，其中有11枚真币和1枚假币。假币和真币的重量不同，但是并不清楚假币比真币轻还是重。现在用一架天平称了这些币3次，称的结果有up，even，down，分别表示天平右边比左边轻，两边同样重，天平右边比左边重。请找出假币并确定其与真币相比是轻是重。
解题思路：对于每一枚硬币先假设他是轻的，看这样是否符合称量结果。如果符合，问题解决。如果不符合，那么就假设他是重的。把所有硬币枚举一遍，一定能找到假币。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>

char left[3][7],right[3][7],result[3][7];

bool isFake(char ch,bool light)
{
for(int i = 0;i < 3;i++){
char *pLeft,*pRight;
if(light){
pLeft = left[i];
pRight = right[i];
}else{
pLeft = right[i];
pRight = left[i];
}

switch(result[i][0]){
case 'u':
if (!strchr(pRight,ch))
return false;
break;
case 'e':
if(strchr(pRight,ch) || strchr(pLeft,ch))
return false;
break;
case 'd':
if(!strchr(pLeft,ch))
return false;
break;
}
}
return true;
}

int main(int argc, char const *argv[])
{
int n;
std::cin >> n;
while(n--){
for(int i = 0;i < 3;i++)
std::cin >> left[i] >> right[i] >> result[i];
for(char ch = 'A';ch <= 'L';++ch){
if(isFake(ch,true)){
std::cout << ch << " is the counterfeit coin and it is light." << std::endl;
}
else if (isFake(ch,false)){
std::cout << ch << " is the counterfeit coin and it is heavy." << std::endl;
}
}
}
return 0;
}