poj 2106

原题链接：http://poj.org/problem?id=2106

题意：或、与、 非的多元表达式的求值；

思路：中缀表达式变为后缀表达式；

代码：

#include<cstdio>
#include<iostream>
#include<stack>
#include<string>

using namespace std;

struct Pri{
    char op;
    int pri;
}lpri[5]={{'(',1},{'|',3},{'&',5},{'!',6},{')',8}},
 rpri[5]={{'(',8},{'|',2},{'&',4},{'!',7},{')',1}};
 
 int find_left(char op)
 {
     for(int i=0;i<5;i++)
         if(lpri[i].op==op)
             return lpri[i].pri;
 }
 
 int find_right(char op)
 {
     for(int i=0;i<5;i++)
         if(rpri[i].op==op)
             return rpri[i].pri;
 }
 
int Precede(char op1,char op2)
{
    int L=find_left(op1);
    int R=find_right(op2);
    if(L==R)    
        return 0;
    if(L>R)
        return 1;
    return -1;     
} 

int Run(stack<int>& num,char op)
{
    int a=num.top();
    num.pop();
    
    if(op=='!')
        return !a;
    int b=num.top();
    num.pop();
    switch(op)
    {
        case '|':
            return a||b;
        case '&':
            return a&&b;
    }
}

int main()
{
    string s;
    int ca=1;
    while(getline(cin,s))
    {
        int i;
        stack<int> num;
        stack<char> op;
         for(i=0;i<s.length();i++)
        {
            if(s[i]==' ')
                continue;    
            else if(s[i]=='V')
                num.push(1);    
            else if(s[i]=='F')
                num.push(0);
            else
            {
                if(op.empty())
                {
                    op.push(s[i]);    
                }
                else
                {
                    int judg=Precede(op.top(),s[i]);
                    int t;
                    switch(judg)
                    {
                        case 0:
                            op.pop();break;
                        case 1:
                            t=Run(num,op.top());
                            num.push(t);
                            op.pop();
                            i--;break;
                        case -1:
                            op.push(s[i]);break;
                    }
                }
            }
        }
        while(!op.empty())
        {
            num.push(Run(num,op.top()));
                op.pop();
        }
        if(num.top()==1)
            printf("Expression %d: V
",ca++);
        else
            printf("Expression %d: F
",ca++);
    }
    
    return 0;
}

注意：特殊数据:!!!!!!!!!!!!!!!F

ac与RE只在一念只差

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