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\n"); break; }
        }

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

    return 0;
}

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