poj 3295 Tautology

来源：互联网发布：材料sci知乎编辑：程序博客网时间：2024/05/21 18:40

链接：http://poj.org/problem?id=3295

模拟了一下：

#include<iostream>#include <stdio.h>#include <algorithm>#include <cmath>#include<stdlib.h>#include <string.h>#include<queue>#include<set>#include<map>#include<stack>#include<time.h>using namespace std;#define MAX_N 105#define inf 0x7fffffff#define LL long long#define ull unsigned long longconst LL INF = 9e18;const int mod = 100000000;typedef pair<double, double>P;char S[MAX_N];stack<char>ope, c;map<char, int>mp;int p, q, r, s, t;int len;void init(){    mp['p'] = p;    mp['q'] = q;    mp['r'] = r;    mp['s'] = s;    mp['t'] = t;}int find(int left, int right){    if(S[left]>='a' && S[left]<='z')        return left;    int sum;    if(S[left] == 'N')        sum = 1;    else        sum = 2;    for(int i=left+1;i<=right;i++) {        if(S[i]>='a' && S[i]<='z')            sum--;        else if(S[i]!='N')            sum++;        if(!sum)            return i;    }    return right;}bool dfs(int left, int right){    if(right == left) {        return mp[S[left]];    }    if(S[left] == 'N') {        return !dfs(left+1, right);    }    else {        if(S[left] <= 'z' && S[left] >= 'a') {            return mp[S[left]];        }        else if(S[left] == 'K') {            int mid = find(left+1, right);            return dfs(left+1, mid) & dfs(mid+1, right);        }        else if(S[left] == 'A') {            int mid = find(left+1, right);            return dfs(left+1, mid) | dfs(mid+1, right);        }        else if(S[left] == 'C') {            int mid = find(left+1, right);            return (!dfs(left+1, mid)) | dfs(mid+1, right);        }        else if(S[left] == 'E') {            int mid = find(left+1, right);            return dfs(left+1, mid) == dfs(mid+1, right);        }    }}bool check(){    for(p=0;p<2;p++) {        for(q=0;q<2;q++) {            for(r=0;r<2;r++) {                for(s=0;s<2;s++) {                    for(t=0;t<2;t++) {                        init();                        if(!dfs(0, len-1)) {                            return false;                        }                    }                }            }        }    }    return true;}int main(){    while(scanf("%s",S)) {        if(S[0] == '0')            break;        len = strlen(S);        if(check())            printf("tautology\n");        else            printf("not\n");    }}

评论区里看到的一个简洁递归：

#include<iostream>#include<cstring>#include<cassert>using namespace std;string s;int cur;char a[5];void Change(char i)//穷举时改变数组用的函数{    for(int j=0;j!=5;j++)        a[j]=((i>>j)&00000001);}bool Cal()//递归{    char c=s[cur];bool w,x;    cur++;    switch(c){    case 'p':return a[0];    case 'q':return a[1];    case 'r':return a[2];    case 's':return a[3];    case 't':return a[4];    case 'N':return !Cal();    case 'K':w=Cal();x=Cal();return (w&&x);    case 'A':w=Cal();x=Cal();return (w||x);    case 'C':w=Cal();x=Cal();return (w<=x);    case 'E':w=Cal();x=Cal();return (w==x);    default :return 0;    }}int main(){    int f;    while(1)    {        cin>>s;        if(s=="0")  break;        f=1;        for(char i=0;i!=32;i++)        {            cur=0;Change(i);            if(!Cal()) {f=0;break;}        }        if(f==1) cout<<"tautology"<<endl;        else cout<<"not"<<endl;    }    return 0;}

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