POJ3295 Tautology(枚举)

题目链接。

分析：

最多有五个变量，所以枚举所有的真假值，从后向前借助于栈验证是否为永真式。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <stack>

using namespace std;

const int maxn = 10000 + 10;

stack<bool> S;
char s[maxn];
bool hash[200], ans;

void check(char pos) {

    if(s[pos] == 'N') {
        bool x = S.top(); S.pop();
        S.push(!x);
    } else if(s[pos] >= 'p' && s[pos] <= 't') {
        S.push(hash[s[pos]]);
    } else {
        bool x, y;
        x = S.top(); S.pop();
        y = S.top(); S.pop();

        if(s[pos] == 'K') S.push(x&y);
        else if(s[pos] == 'A') S.push(x|y);
        else if(s[pos] == 'C') S.push(!x | y); //即蕴含
        else if(s[pos] == 'E') S.push(!(x^y));
    }

}

int main() {
    //freopen("my.txt", "r", stdin);
    while(scanf("%s", s) == 1) {
        if(s[0] == '0') break;
        ans = true;
        int i;

        for(i=0; i<(1<<5); i++) {
            hash['p'] = i & 1;
            hash['q'] = i & (1<<1);
            hash['r'] = i & (1<<2);
            hash['s'] = i & (1<<3);
            hash['t'] = i & (1<<4);

            int pos = strlen(s)-1;

            while(pos >= 0) {
                check(pos);
                pos--;
            }

            ans = S.top(); S.pop();

            if(ans == false) { printf("not
"); break; }
        }

        if(i >= (1<<5)) printf("tautology
");
    }

    return 0;
}

另一种写法

该写法参考自Disuss：http://poj.org/showmessage?message_id=168123（里面对于莫名其妙的错误还有讲解，不错）

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <stack>

using namespace std;

const int maxn = 10000 + 10;

char s[maxn];
int c_s, value;

bool getval() {
    bool t1, t2;

    switch(s[c_s++]) {
    case 'p': return (value & (1<<0));
    case 'q': return (value & (1<<1));
    case 'r': return (value & (1<<2));
    case 's': return (value & (1<<3));
    case 't': return (value & (1<<4));

    case 'N': return !getval();
    case 'K': t1 = getval(); t2 = getval(); return t1 & t2;
    case 'A': t1 = getval(); t2 = getval(); return t1 | t2;
    case 'C': t1 = getval(); t2 = getval(); return (!t1 | t2);
    case 'E': t1 = getval(); t2 = getval(); return (!(t1^t2));
    }
}

int main() {
//  freopen("my.txt", "r", stdin);
    while(scanf("%s", s) == 1) {
        if(s[0] == '0') break;

        for(value = 0; value < (1<<5); value++){
            c_s = 0;
            if(!getval()) { printf("not
"); break; }
        }

        if(value >= (1<<5)) printf("tautology
");
    }

    return 0;
}