POJ1013 Counterfeit Dollar

POJ1013 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个硬币，其中一个为假币，11真币，假币不知轻重，现给出三次称量方式（左右盘各有哪几枚硬币）及结果（右边盘子高或低或平），问哪一个是假币，轻还是重。

算法

穷举暴力，每个Case中穷举假设12 * 2 次（12个中某个硬币假设为假币，有两种轻重可能）检查一遍，看到和称量结果一致的假设就输出。

AC代码

#include<stdio.h>
#include<string.h>

typedef enum {left = -1, even = 0, right = 1} weigh;    
//Which side is higher?
typedef enum {light = 0, heavy = 1} Fake;               
//Assumption: fake coin---light or heavy?

typedef struct {
    char Left[10];
    char Right[10];
    weigh Status;
} Balance;                                              
//one balance --- left side right side and status

void GetWeighing(Balance Weighing[]);                   
//Get the weighing result
int TestWeigh(Balance Weighing[], char coin, Fake how); 
//Is our assumption reasonable?
weigh Judge(char * stat);

int main()
{   
    Balance Weighing[3];
    char *Counterfeit[2] = {"light", "heavy"};
    int N;

    scanf("%d", &N);
    for (int i = 0; i < N; i++) {
        GetWeighing(Weighing);
        for (int j = 0; j < 12; j++) {                  
        //test 12 * 2 assumptions
            Fake how = light;
            if (TestWeigh(Weighing, 'A' + j, how)) {
                printf("%c is the counterfeit coin and it is %s.\n", 'A' + j, Counterfeit[(int)how]);
            }
            else {
                how = heavy;
                if (TestWeigh(Weighing, 'A' + j, how)) {
                    printf("%c is the counterfeit coin and it is %s.\n", 'A' + j, Counterfeit[(int)how]);
                }
            }
        }
    }

    return 0;
}

void GetWeighing(Balance Weighing[])
{
    char stat[5];

    for (int i = 0; i < 3; i++) {
        scanf("%s %s", Weighing[i].Left, Weighing[i].Right);
        scanf("%s", stat);
        Weighing[i].Status = Judge(stat);
    }
}

weigh Judge(char * stat)
{
    if (stat[0] == 'e') {
        return even;
    }
    else if (stat[0] == 'u') {
        return right;
    }
    else {
        return left;
    }
}

int TestWeigh(Balance Weighing[], char coin, Fake how)
{
    weigh Shouldbe;
    //result according to the assumption
    int BOOL = 1;

    for (int k = 0; k < 3; k++) {
        char * l = strchr(Weighing[k].Left, coin), * r = strchr(Weighing[k].Right, coin);
        if (l == NULL && r == NULL) {   
            Shouldbe = even;
        }
        else if ((l == NULL && how == light) || (r == NULL && how == heavy)) {
            Shouldbe = right;
        }
        else {
            Shouldbe = left;
        }
        if (Shouldbe != Weighing[k].Status) {
            BOOL = 0;
        }
    }

    return BOOL;
}