C++实现求主析取范式、主合取范式

离散数学第一次上机实验，突击学完什么是主析取范式和主合取范式之后粗略地想了下实现的思路。

第一步自然是把输入的式子转换成后缀表达式。结合数据结构书上对基本四则运算的代码和前人的思路勉强写出来一个，但还是有很多地方可能一转头就会忘了怎么实现。

转换成后缀表达式之后的事情就显得轻松多了。

下面贴代码。

#include<iostream>#include<string>using namespace std;class SeqStack {public:SeqStack(int mSize);~SeqStack();bool Push(char x);bool Pop();char Top();bool Empty();private:int top;char *stack;int maxtop;};SeqStack::SeqStack(int mSize) {maxtop = mSize - 1;stack = new char[mSize];top = -1;}SeqStack::~SeqStack() {delete[]stack;}bool SeqStack::Push(char x) {if (top == maxtop)return false;stack[++top] = x;return true;}bool SeqStack::Pop() {if (-1==top )return false;top--;return true;}bool SeqStack::Empty() {if (-1==top)return true;return false;}char SeqStack::Top() {return stack[top];}bool IsLegal(const string orgin) {for (unsigned int i = 0; i < orgin.length(); i++) {if (orgin[i] != 'P' && orgin[i] != 'Q' && orgin[i] != 'R' && orgin[i] != '&' && orgin[i] != '|' && orgin[i] != '=' && orgin[i] != '>'&&orgin[i]!='!')return false;}return true;}int And(int a,int b, SeqStack &stack) { int res = a * b;stack.Push(res);return res;}int Or (int a, int b, SeqStack &stack) {int res = a + b;res = res > 1 ? 1 : res;stack.Push(res);return res;}int Not(int a, SeqStack &stack) {  a = stack.Top();stack.Pop();int res = a == 1 ? 0 : 1;stack.Push(res);return res;}int If(int a, int b, SeqStack &stack) { //条件int res = (b == 1 && a == 0) ? 0 : 1;stack.Push(res);return res;}int Iif(int a, int b, SeqStack &stack) {//双条件int res = (b == a) ? 1 : 0;stack.Push(res);return res;}bool CanIn(char out,SeqStack &stack) {        //先计算优先级，然后判断能否进栈char in = stack.Top();int i=0, o=0;  //分别表示栈内外运算符的优先级switch (in) {case '#':i = 0; break;case '(':i = 1; break;case '=':i = 3; break;case '>':i = 5; break;case '|':i = 7; break;case '&':i = 9; break;case '!':i = 11; break;case ')':i = 12; break;}switch (out) {case '#':o = 0; break;case '(':o = 12; break;case '=':o = 2; break;case '>':o = 4; break;case '|':o = 6; break;case '&':o = 8; break;case '!':o = 10; break;case ')':o = 1; break;}if (i < o)      //栈外优先级比栈内高就进栈return true;elsereturn false;}void InfixToPostfix(string &orgin ,SeqStack &stack) {  //转换为后缀表达式string tmp;                //保存后缀表达式stack.Push('#');                for (int i = 0; (unsigned)i <orgin.length(); i++) {if (orgin[i] == 'P' || orgin[i] == 'Q' || orgin[i] == 'R' ) {tmp = tmp + orgin[i];continue;}if (CanIn(orgin[i],stack))stack.Push(orgin[i]);else if (orgin[i] == ')'){while (stack.Top() != '('){tmp = tmp + stack.Top();stack.Pop();}stack.Pop();}else{do{tmp = tmp + stack.Top();stack.Pop();} while (!CanIn(orgin[i],stack));stack.Push(orgin[i]);}}while (stack.Top() != '#'){tmp = tmp + stack.Top();stack.Pop();}stack.Pop();                // '#' 出栈orgin = tmp;}int Calculate(int number,string &org,SeqStack &stack, int p, int q, int r){int value=0,a=0,b=0,res=0;if (number == 3){for (int i = 0; (unsigned)i <org.length(); i++){if (org[i] == 'P' || org[i] == 'Q' || org[i] == 'R'){value = org[i] == 'P' ? p : org[i] == 'Q' ? q : r;stack.Push(value);continue;}if (org[i] != '!'){a = stack.Top();stack.Pop();b = stack.Top();stack.Pop();}switch (org[i]){case '=':res=Iif(a,b,stack); break;case '>':res=If(a, b, stack); break;case '|':res=Or(a, b, stack); break;case '&':res=And(a, b, stack); break;case '!':res=Not(a, stack); break;}}}if (number == 2){for (int i = 0;(unsigned)i <org.length(); i++){if (org[i] == 'P' || org[i] == 'Q'){value = org[i] == 'P' ? p : q;stack.Push(value);continue;}if (org[i] != '!'){a = stack.Top();stack.Pop();b = stack.Top();stack.Pop();}switch (org[i]) {case '=':res = Iif(a, b, stack); break;case '>':res = If(a, b, stack); break;case '|':res = Or(a, b, stack); break;case '&':res = And(a, b, stack); break;case '!':res = Not(a, stack); break;}}}return res;}char num2bool(int t) {if (0 == t)return 'T';else return 'F';}void Print(int number, string &orgin, SeqStack &stack) {int p, q, r, res;string pcnf;    //主合取范式string  pdnf;    //主析取范式if (number == 3) {cout << "P\t" << "Q\t" << "R\t" << "Z" << endl;for (p = 1; p >= 0; p--) {for (q = 1; q >= 0; q--) {for (r = 1; r >= 0; r--) {res = Calculate(3, orgin, stack, p, q, r);if (res == 1)pdnf = pdnf + "(" + (p == 1 ? "P" : "!P") + "&" + (q == 1 ? "Q" : "!Q") + "&"+ (r == 1 ? "R" : "!R") + ")" + " | ";elsepcnf = pcnf + "(" + (p == 0 ? "P" : "!P") + "|" + (q == 0 ? "Q" : "!Q") + "|"+ (r == 0 ? "R" : "!R") + ")" + " & ";char cp, cq, cr, cres;cp = num2bool(p);cq = num2bool(q);cr = num2bool(r);cres = num2bool(res);cout << cp << "\t" << cq << "\t" << cr << "\t" << cres << endl;}}}}if (number == 2) {cout << "P\t" << "Q\t" << "Z" << endl;for (p = 1; p >= 0; p--) {for (q = 1; q >= 0; q--) {res=Calculate(2, orgin, stack, p, q, 0);if (res == 1)pdnf = pdnf + "(" + (p == 1 ? "P" : "!P") + "&" + (q == 1 ? "Q" : "!Q") + ")" + " | ";elsepcnf = pcnf + "(" + (p == 0 ? "P" : "!P") + "|" + (q == 0 ? "Q" : "!Q") + ")" + " & ";char cp, cq, cres;cp = num2bool(p);cq = num2bool(q);cres = num2bool(res);cout << cp << "\t" << cq << "\t" << cres << endl;}}}//永真和永假的情况下长度为0if (pdnf.length() != 0)pdnf.erase(pdnf.length() - 2);if (pcnf.length() != 0)pcnf.erase(pcnf.length() - 2);cout << "主析取范式：" << pdnf << endl << endl;cout << "主合取范式：" << pcnf << endl << endl;}/****************************主函数******************************/int main() {SeqStack stack(200);int cnt = 0;string orgin;int number;cout << "/**************输入提示*****************/" << endl;cout << "/*              ! 否定                 */" << endl;cout << "/*              | 析取                 */" << endl;cout << "/*              & 合取                 */" << endl;cout << "/*              > 条件                 */" << endl;cout << "/*              = 双条件               */" << endl;cout << "/**************输入提示*****************/" << endl;cout << "请输入变元数量 (2 或 3)" << endl;loop:cin >> number;if(2==number) cout << endl << "变元请用P、Q表示" << endl ;else if (3==number)cout << endl << "变元请用P、Q、R表示" << endl;else {cout << "请输入2或3：  " << endl;goto loop;}cout << "输入命题：";int flag = 0;do {if (flag != 0) {cout << "输入包含不合法字符，请仔细阅读输入提示并重新输入\n（符号请使用半角英文符号）" << endl;cout << "输入命题：";}cin >> orgin;flag = 1;} while (!IsLegal(orgin));InfixToPostfix(orgin,stack);Print(number,orgin,stack);while (1);return 0;}