POJ1013Counterfeit Dollar

Counterfeit Dollar

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 43484		Accepted: 13785

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. 

借鉴一下别人写的很详细的讲解

大致题意：

有一打（12枚）硬币，其中有且仅有1枚假币，11枚真币

用A~L作为各个硬币的代号

假币可能比真币略轻，也可能略重

现在利用天枰，根据Input输入的3次称量，找出假币，并输出假币是轻还是重。

 

解题思路：

模拟法要考虑的情况较繁琐，可利用简单的逻辑推理进行解题。

 

 

注意Input一行代表一次称量，每行有三个字符串，分别为

Left   right     status

代表该次称量时，天枰左盘放的硬币、天枰右盘放的硬币、天枰右盘的状态

 

共三种状态：

Up：右盘上升，说明右盘可能有轻假币，也可能左盘有重假币。

Down：右盘下降，说明右盘可能有重假币，也可能左盘有轻假币。

Even：右盘与左盘平衡，由于假币有且仅有1枚，则说明此时天枰两边的硬币全为真币。

 

注意题目的字眼：

1、  有且仅有1枚假币

2、  假币相对于真币的重量，可能轻可能重

3、  只称量3次，且称量3次恰好且必能找到假币

4、  每次称量时天枰两边的硬币数目一样

5、  选取哪些硬币称量由input决定

 

从3、4、5可知，由于无法知道每次选取称量的硬币，那么3次称量可能只选用了几个硬币，也可能仅有一两个硬币没有选上，那么用模拟法去记录每次用于称量的硬币的状态（真假，其中假币又有轻重之分）并推导没有被称量的硬币状态（或状态变化）是很困难的，虽然人很容易做到这点，但计算机却很难去“推导”，因为称量硬币的方法是无规律的且非常多。

 

那么只能通过适当转化问题后用另一种有效的方法去解决。

 

虽然称量硬币的方法是无规律且未知的，但是称量硬币后的结果却只有3个，up、down和 even。且当出现even时，天枰两边的硬币必然都为真币，假币必定在余下的硬币之间（这是因为假币有且只有一枚），那么我们就可以定义一个标记数组 zero[]去标记even时的真币，在以后的处理把他们排除在外。

而唯一难以处理的是up和down的状态，因为假币可能轻可能重，则这两种状态都无法得知究竟假币出现在天枰的哪边。

 

处理up和down状态方法：

当出现up或down状态时，天枰两边的所有硬币都应该被怀疑为假币（已标记必定为真币的硬币不必被怀疑）。

首先time[]记录每个硬币的被怀疑程度，time[i]=0表示该硬币i不被怀疑（即其可能为真币）。定义在up状态盘的硬币为“轻怀疑假币”，通过“--”操作加深其被怀疑为轻假币的程度，“负号”为轻假币的怀疑方向；在down状态盘的硬币为“重怀疑假币”，通过“++”操作加深其被怀疑为重假币的程度，“正号”为重假币的怀疑方向。

那么若一枚真币被怀疑为“轻假币”时，它就可能通过下次称量通过“++”操作取消嫌疑了。初始化所有硬币的怀疑程度均为0。

称量完毕后，找出被怀疑程度最大（注意取绝对值）的硬币，它就是假币。而当其怀疑方向为正时，则其为重假币。为负时，为轻假币。

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;

char s1[20],s2[20],s[10];
int check['L'+1],time['L'+1];

int main()
{
    int T;
    scanf("%d",&T);

    while(T--)
    {
       // getchar();
        memset(check,0,sizeof check);
        memset(time,0,sizeof time);
        for(int j=0;j<3;j++)
        {
            scanf("%s%s%s",s1,s2,s);
            //cout<<s1<<"     "<<s2<<"    "<<s<<"     "<<endl<<"********"<<endl;

            int len1=strlen(s1);
            if(s[0]=='e')
            {
                for(int i=0;i<len1;i++)
                {
                    check[s1[i]]=1;
                    check[s2[i]]=1;
                }
            }
            else if(s[0]=='u')
            {
                for(int i=0;i<len1;i++)
                {
                    time[s1[i]]++;
                    time[s2[i]]--;
                }
            }
            else if(s[0]=='d')
            {
                for(int i=0;i<len1;i++)
                {
                    time[s1[i]]--;
                    time[s2[i]]++;
                }
            }
        }

        char ans;
        int mmax=-1;
        for(char i='A';i<='L';i++)
        {
            if(check[i]) continue;
            if(mmax<=abs(time[i]))
            {
                mmax=abs(time[i]);
                ans=i;
            }
        }
        cout<<ans;
        printf(" is the counterfeit coin and it is ");
        if(time[ans]>0)
            cout<<"heavy.";
        else
            cout<<"light.";
        puts("");

    }
    return 0;
}