Poj 1013 Counterfeit Dollar【暴力枚举】

Counterfeit Dollar

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 44613		Accepted: 14135

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.

Source

East Central North America 1998

题目大意：

有t组数据；

保证一共只有12个字母，每个字母代表一个钞票，钞票有可能轻，也有可能重，给你三个语句，让你输出那个假钞的信息，保证有解。

思路:

1、最不容易错的方法也就是暴力枚举。先将Even语句中的所有钞票定义为不可能是假钞，然后我们再对其他钞票（当然是在语句中出现过才行）的钞票进行以下方式枚举：

①首先将所有钞票的重量定为2

②然后枚举当前可能是假钞的钞票的重量，要么可能是1，要么可能是3（要么轻一点，要么重一点）

③然后将三个语句进行判断，如果能够保证三个语句都符合条件，那么这个钞票就是假的，那么定义出来的值如果是1当然就是假钞轻一些，反之则重一些。

2、注意不要枚举出没有在语句中出现过的字母。

Ac代码:

#include<stdio.h>
#include<string.h>
using namespace std;
char a[3][50];
char b[3][50];
char c[3][50];
int val[15];
int vis[15];
int vis2[15];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(vis,0,sizeof(vis));
memset(vis2,0,sizeof(vis2));
for(int i=1; i<=12; i++)
{
val[i]=2;
}
for(int i=0; i<3; i++)
{
scanf("%s%s%s",a[i],b[i],c[i]);
for(int k=0; k<strlen(a[i]); k++)
{
vis2[a[i][k]-'A'+1]=1;
}
for(int k=0; k<strlen(b[i]); k++)
{
vis2[b[i][k]-'A'+1]=1;
}
if(strcmp(c[i],"even")==0)
{
for(int k=0; k<strlen(a[i]); k++)
{
vis[a[i][k]-'A'+1]=1;
}
for(int k=0; k<strlen(b[i]); k++)
{
vis[b[i][k]-'A'+1]=1;
}
}
}
for(int i=1; i<=12; i++)
{
if(vis[i]==1)continue;
if(vis2[i]==0)continue;
val[i]=1;
int flag=1;
for(int j=0; j<3; j++)
{
int sum1=0;
for(int k=0; k<strlen(a[j]); k++)
{
sum1+=val[a[j][k]-'A'+1];
}
int sum2=0;
for(int k=0; k<strlen(b[j]); k++)
{
sum2+=val[b[j][k]-'A'+1];
}
if(strcmp(c[j],"even")==0)
{
if(sum1!=sum2)
{
flag=0;
}
}
if(strcmp(c[j],"up")==0)
{
if(sum1<=sum2)
{
flag=0;
}
}
if(strcmp(c[j],"down")==0)
{
if(sum1>=sum2)
{
flag=0;
}
}
}
if(flag==1)
{
if(val[i]<2)
{
printf("%c is the counterfeit coin and it is light.\n",i+'A'-1);
}
else printf("%c is the counterfeit coin and it is heavy.\n",i+'A'-1);
}
val[i]=3;
flag=1;
for(int j=0; j<3; j++)
{
int sum1=0;
for(int k=0; k<strlen(a[j]); k++)
{
sum1+=val[a[j][k]-'A'+1];
}
int sum2=0;
for(int k=0; k<strlen(b[j]); k++)
{
sum2+=val[b[j][k]-'A'+1];
}
if(strcmp(c[j],"even")==0)
{
if(sum1!=sum2)flag=0;
}
if(strcmp(c[j],"up")==0)
{
if(sum1<=sum2)flag=0;
}
if(strcmp(c[j],"down")==0)
{
if(sum1>=sum2)flag=0;
}
}
if(flag==1)
{
if(val[i]<2)
{
printf("%c is the counterfeit coin and it is light.\n",i+'A'-1);
}
else printf("%c is the counterfeit coin and it is heavy.\n",i+'A'-1);
}
val[i]=2;
}
}
}