poj3295 Tautology 构造 数据结构 枚举

Tautology

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12321 Accepted: 4695

Description

WFF 'N PROOF is a logic game played with dice. Each die has six faces representing some subset of the possible symbols K, A, N, C, E, p, q, r, s, t. A Well-formed formula (WFF) is any string of these symbols obeying the following rules:

• p, q, r, s, and t are WFFs
• if w is a WFF, Nw is a WFF
• if w and x are WFFs, Kwx, Awx, Cwx, and Ewx are WFFs.

The meaning of a WFF is defined as follows:

• p, q, r, s, and t are logical variables that may take on the value 0 (false) or 1 (true).
• K, A, N, C, E mean and, or, not, implies, and equals as defined in the truth table below.

Definitions of K, A, N, C, and E     w  x  Kwx  Awx   Nw  Cwx  Ewx  1  1  1  1   0  1  1  1  0  0  1   0  0  0  0  1  0  1   1  1  0  0  0  0  0   1  1  1

A tautology is a WFF that has value 1 (true) regardless of the values of its variables. For example, ApNp is a tautology because it is true regardless of the value of p. On the other hand, ApNq is not, because it has the value 0 for p=0, q=1.

You must determine whether or not a WFF is a tautology.

Input

Input consists of several test cases. Each test case is a single line containing a WFF with no more than 100 symbols. A line containing 0 follows the last case.

Output

For each test case, output a line containing tautology or not as appropriate.

Sample Input

ApNpApNq0

Sample Output

tautologynot

Source

Waterloo Local Contest, 2006.9.30

题意：

有p、q、r、s、t五个自变量（取0或1），根据题目给的K, A, N, C, E是作用于自变量的操作。操作结果通过查题目给的真值表得出（好离散啊。。）

现在输入多行数据，让我们判断这个句子是不是永真式

这题的分类是构造，然而并不是很懂构造是个啥玩意儿

抱着学习构造的方法差了这题。。。。不小心瞄到用stack来处理就秒懂了。。。。

有种被剧透的感觉。。。。要多练习啊。。

对输入的句子从后往前操作，遇到小写字母将它如栈，大写字母取头两个或一个出栈操作，然后将计算结果入栈

对pqrst枚举所有可能性，因为是2^5所以数据量很小

不过还是在枚举的时候遇到问题，看来我还是递归用的不是很熟练，这个也是个记录路径的问题

代码：

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cmath>#include <stack>using namespace std;int cha[60][5];stack<int> judge;int num,len;string in;void init(int index){    int i;    if(index>4){        num++;        for(i=0;i<=4;i++){            cha[num][i] = cha[num-1][i];        }        return;    }        cha[num][index] = 1;    init(index+1);    cha[num][index] = 0;    init(index+1);}bool run(int index){    int i,j;    while(judge.size()){        judge.pop();    }    int w,x,res;    for(i=len-1;i>=0;i--){                if((int)in[i]<112){            if(in[i]=='K'){                w = judge.top();                judge.pop();                x = judge.top();                judge.pop();                if(w&&x){                    res = 1;                }else{                    res = 0;                }                judge.push(res);            }else if(in[i]=='A'){                w = judge.top();                judge.pop();                x = judge.top();                judge.pop();                if(w==0&&x==0){                    res = 0;                }else{                    res = 1;                }                judge.push(res);            }else if(in[i]=='N'){                w = judge.top();                judge.pop();                res = !w;                judge.push(res);            }else if(in[i]=='C'){                w = judge.top();                judge.pop();                x = judge.top();                judge.pop();                if(w==1&&x==0){                    res = 0;                }else{                    res = 1;                }                judge.push(res);            }else{//'E'                w = judge.top();                judge.pop();                x = judge.top();                judge.pop();                if(w==x){                    res = 1;                }else{                    res = 0;                }                judge.push(res);            }        }else{            judge.push(cha[index][in[i]-112]);        }    }    if(judge.top()){        return true;    }else{        return false;    }}int main(){    //int test = 'p'; //K is 75    //cout<<test<<endl;// -->r is 112    int i,j,k;    init(0);    /*for(i=0;i<=32;i++){        for(j=0;j<5;j++){            cout<<cha[i][j]<<" ";        }        cout<<endl;    }*/            while(cin>>in){        if(in=="0"){            break;        }        len = in.length();        num = 0;        //cout<<"num is "<<num<<endl;-->32        bool flag = true;        for(i=0;i<=32;i++){            if(!run(i)){                flag = false;            }        }        if(flag){            printf("tautology\n");        }else{            printf("not\n");        }    }        return 0;}