CSUOJ 1637 Yet Satisfiability Again!

1637: Yet Satisfiability Again!

Time Limit: 5 Sec  Memory Limit: 128 MB

Description

Alice recently started to work for a hardware design company and as a part of her job, she needs to identify defects in fabricated ntegrated circuits. An approach for identifying these defects boils down to solving a satisfiability instance. She needs your help to write a program to do this task.

Input

The first line of input contains a single integer, not more than 5, indicating the number of test cases to follow. The first line of each test case contains two integers n and m where 1 ≤ n ≤ 20 indicates he number of variables and 1 ≤ m ≤ 100 indicates the number of clauses. Then, m lines follow corresponding to each clause. Each clause is a disjunction of literals in the form Xi or ~Xi for some 1 ≤ i ≤ n, where Xi indicates the negation of the literal Xi. The “or” operator is denoted by a ‘v’ character and is seperated from literals with a single pace.

Output

For each test case, display satisfiable on a single line if there is a satisfiable assignment; otherwise
display unsatisfiable.

Sample Input

2
3 3
X1 v X2
~X1
~X2 v X3
3 5
X1 v X2 v X3
X1 v ~X2
X2 v ~X3
X3 v ~X1
~X1 v ~X2 v ~X3

Sample Output

satisfiable
unsatisfiable

HINT

 

Source

解题:以为可以用位运算优化，结果，却发现，错了！好吧，还是上暴力的吧

 #include <bits/stdc++.h>
 #define pii pair<int,int>
 using namespace std;
 const int maxn = ;
 char str[maxn];
 vector< pii >s[maxn];
 int n,m;
 void exp2num(int idx) {
     bool flag = true;
     for(int i = ,ret = ; str[i];) {
         if(str[i] == '~') flag = false;
         if(isdigit(str[i])) {
             while(isdigit(str[i])) ret = ret* + (str[i++]-'');
             s[idx].push_back(make_pair(ret-,flag));
             ret = ;
             flag = true;
         } else i++;
     }
 }
 bool check(int o){
     for(int i = ; i < m; ++i){
         bool flag = false;
         for(int j = s[i].size()-; j >= ; --j){
             if(s[i][j].second == ((o>>s[i][j].first)&)){
                 flag = true;
                 break;
             }
         }
         if(!flag) return false;
     }
     return true;
 }
 int main(){
     int T;
     scanf("%d",&T);
     while(T--){
         scanf("%d %d",&n,&m);
         getchar();
         for(int i = ; i < m; ++i){
             s[i].clear();
             gets(str);
             exp2num(i);
         }
         bool ok = false;
         for(int i = ; i < (<<n) && !ok; ++i)
             ok = check(i);
         puts(ok?"satisfiable":"unsatisfiable");
     }
     return ;
 }