首页 > ACM题库 > HDU-杭电 > HDU 1482 Counterfeit Dollar-模拟-[解题报告] C++
2013
12-11

HDU 1482 Counterfeit Dollar-模拟-[解题报告] C++

Counterfeit Dollar

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

1
ABCD EFGH even
ABCI EFJK up
ABIJ EFGH even

样例输出:

K is the counterfeit coin and it is light.

有点小麻烦的模拟。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<map>
#include<string>
#include<cstring>
#include<cmath>
using namespace std;

int vis['Z'];
int istrue['Z'];
int main()
{
	int t,i,j;
	scanf("%d",&t);
	while(t--)
	{
		int min=0x3f3f3f3f;
		int max=-0x3f3f3f3f;
		int eps=max;								//计算与特殊值的差					
		string a[3],b[3],c[3];
		memset(vis,0,sizeof(vis));
		memset(istrue,0,sizeof(istrue));
		cin>>a[0]>>b[0]>>c[0];
		cin>>a[1]>>b[1]>>c[1];
		cin>>a[2]>>b[2]>>c[2];
		for(i=0;a[0][i];i++)
			vis[a[0][i]]=vis[b[0][i]]=1000;
		for(i=0;a[1][i];i++)
			vis[a[1][i]]=vis[b[1][i]]=1000;
		for(i=0;a[2][i];i++)
			vis[a[2][i]]=vis[b[2][i]]=1000;			//给出现过的字符标上特殊值。
		for(i=0;i<3;i++)
		{
			if(c[i]=="even") 
			{
				for(j=0;a[i][j];j++)
				{
					istrue[a[i][j]]=1;				//istrue数组记录绝对正确的硬币
					istrue[b[i][j]]=1;
				}
			}
			if(c[i]=="up")
			{
				for(j=0;a[i][j];j++)
				{
					vis[ a[i][j] ]++;
					vis[ b[i][j] ]--;
				}
			}
			if(c[i]=="down")
			{
				for(j=0;a[i][j];j++)
				{
					vis[ a[i][j] ]--;
					vis[ b[i][j] ]++;
				}
			}
		}
		int ok,k;
		for(i='A';i<='L';i++)
		{
			ok=1;
			if(vis[i]==0) continue;
			if(vis[i]<min) min=vis[i];
			if(vis[i]>max) max=vis[i];
			if(istrue[i]) continue;
			
			for(j='A';j<='L';j++)						//第一次判断,假币在++,--,计算后很可能是一个只出现一次的值
			{
				if(i==j||vis[j]==0||istrue[j]) continue;
				if(vis[i]==vis[j])
				{ok=0;break;}
			}
			if(ok)
			{
				
				if(abs(vis[i]-1000)>eps) {eps=abs(vis[i]-1000);k=i;}	//再一次判断,如果相差越大,说明越是假币				
			}
		}
		if(vis[k]==min) printf("%c is the counterfeit coin and it is light.\n",k);
		else printf("%c is the counterfeit coin and it is heavy.\n",k);
	}
	return 0;
}
/*
送数据
sample input 
12 
ABCD EFGH even 
ABCI EFJK up 
ABIJ EFGH even 
AGHL BDEC even 
JKI ADE up 
J K even 
ABCDEF GHIJKL up 
ABC DEF even 
I J down 
ABCDEF GHIJKL up 
ABHLEF GDIJKC down 
CD HA even 
A B up 
B A down 
A C even 
A B up 
B C even 
DEFG HIJL even 
ABC DEJ down 
ACH IEF down 
AHK IDJ down 
ABCD EFGH even 
AB IJ even 
A L down 
EFA BGH down 
EFC GHD even 
BA EF down 
A B up 
A C up 
L K even 
ACEGIK BDFHJL up 
ACEGIL BDFHJK down 
ACEGLK BDFHJI down 
ACEGIK BDFHJL up 
ACEGIL BDFHJK down 
ACEGLK BDFHJI up 

sample output 
K is the counterfeit coin and it is light. 
I is the counterfeit coin and it is heavy. 
I is the counterfeit coin and it is light. 
L is the counterfeit coin and it is light. 
B is the counterfeit coin and it is light. 
A is the counterfeit coin and it is heavy. 
A is the counterfeit coin and it is light. 
L is the counterfeit coin and it is heavy. 
A is the counterfeit coin and it is light. 
A is the counterfeit coin and it is heavy. 
L is the counterfeit coin and it is light. 
K is the counterfeit coin and it is heavy.
*/

解题报告转自:http://blog.csdn.net/t1019256391/article/details/8928422