poj 3295 Tautology(经典构造算法题)

 

思路：1）使用递归模拟，用备忘录优化，否则超时

                另外：学到了一个不用递归即可枚举构造0-1序列的方法

for(i=0;i<32;i++)
	for(j=0;j<5;j++)
		arr[j]=(i>>j)%2;

 

【源程序】：

#include "stdio.h"
#include "string.h"

int arr[101],flag,note[101][101];

int IsCorrect(char *s,int start,int end)
{
	if(note[start][end]!=-1) return note[start][end];

	int i;
	if(end==start)
	{
		if(s[start]<='t'&&s[start]>='p')
		{
			note[start][end]=arr[s[start]-'p'];
			return note[start][end];
		}
		else
		{
			note[start][end]=2;
			return note[start][end];
		}
	}
	else if(s[start]=='N')
	{
		if(IsCorrect(s,start+1,end)==0)
		{
			note[start][end]=1;
			return note[start][end];
		}
		else if(IsCorrect(s,start+1,end)==1)
		{
			note[start][end]=0;
			return note[start][end];
		}
		else
		{
			note[start][end]=2;
			return note[start][end];
		}
	}
	else if(s[start]=='E'||s[start]=='C'||s[start]=='A'||s[start]=='K')
	{
		int sig=2;
		for(i=1;i<end-start;i++)
		{
			int a=IsCorrect(s,start+1,start+i);note[start+1][start+i]=a;
			int b=IsCorrect(s,start+i+1,end);note[start+i+1][end]=b;
			if(a<=1 && b<=1)
			{
				switch(s[start])
				{
					case'K':sig=(a&&b);break;   
				 		 
				  	case'A':sig=(a||b);break;      
	
				  	case'C':sig=(!a||b);break;       
	
				  	case'E':sig=!(a^b);break;         
				}
			}
			if(sig==1)
				return note[start][end]=1;
		}
		return note[start][end]=sig;
	}
	else
	{
		return (note[start][end]=2);
	}
}
int main()
{
	//freopen("input.txt","r",stdin);
	int len,i,j;
	char s[260];
	while((scanf("%s",s))!=EOF)
	{
		if(strcmp(s,"0")==0) break;
		len=strlen(s);
		memset(arr,0,sizeof(arr));
		flag=1;
		for(i=0;i<32;i++)
		{
			for(j=0;j<5;j++)
				arr[j]=(i>>j)%2;
			
			for(int p=0;p<len;p++)
			    for(int q=0;q<len;q++)
			        note[p][q]=-1;
			flag=IsCorrect(s,0,len-1);
			if(flag!=1) break;
		}
		if(flag==1)
			printf("tautology\n");
		else
			printf("not\n");
	}
	return 0;
}

 

           2）别人的思路：用栈模拟0MS

【程序】：

#include<iostream>
#include<string>
#include<stack>
using namespace std;
bool Judge(int p,int q,int r,int s,int t,string st)
{
	stack<int> s_num;
	int buff_1,buff_2;
	string::size_type n = st.size();
	for(n=n-1;n!=-1;n--)
	{
		if(st[n]<='t'&&st[n]>='p')
		{
			switch(st[n])
			{
				case 't':s_num.push(t);continue;
				case 's':s_num.push(s);continue;
				case 'r':s_num.push(r);continue;
				case 'q':s_num.push(q);continue;
				case 'p':s_num.push(p);continue;
			}
		}
		else
		{
			if(st[n]=='N')
			{
				buff_1 = s_num.top();s_num.pop();
				s_num.push(!buff_1);
			}
			else	if(st[n]=='A')
			{
				buff_1 = s_num.top();s_num.pop();
				buff_2 = s_num.top();s_num.pop();
				s_num.push(buff_2||buff_1);
			}
			else	if(st[n]=='K')
			{
				buff_1 = s_num.top();s_num.pop();
				buff_2 = s_num.top();s_num.pop();
				s_num.push(buff_2&&buff_1);
			}
			else	if(st[n]=='E')
			{
				buff_1 = s_num.top();s_num.pop();
				buff_2 = s_num.top();s_num.pop();
				s_num.push(buff_2==buff_1);
			}
			else
			{
				buff_1 = s_num.top();s_num.pop();
				buff_2 = s_num.top();s_num.pop();
				s_num.push((!buff_1)||buff_2);
			}
		}
	}
	//cout<<s_num.top()<<endl;
	if(!s_num.top())
		return false;
	return true;
}
int main()
{
	string st;
	while(cin>>st)
	{
		string::size_type i = 0;
		if(st[i]=='0')
			break;
		int p(0),q(0),r(0),s(0),t(0),flag(0);
		for(p=0;p!=2;p++)
		{
			for(q=0;q!=2;q++)
			{
				for(r=0;r!=2;r++)
				{
					for(s=0;s!=2;s++)
					{
						for(t=0;t!=2;t++)
						{
							if(!Judge(p,q,r,s,t,st))
							{
								flag=1;
							}
						}
						if(flag==1)
							break;
					}
					if(flag==1)
						break;
				}
				if(flag==1)
					break;
			}
			if(flag==1)
				break;
		}
		if(flag==1)
			cout<<"not"<<endl;
		else
			cout<<"tautology"<<endl;
	}
	return 0;
}