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

 1 #include <bits/stdc++.h>

 2 #define pii pair<int,int>

 3 using namespace std;

 4 const int maxn = 200;

 5 char str[maxn];

 6 vector< pii >s[maxn];

 7 int n,m;

 8 void exp2num(int idx) {

 9     bool flag = true;

10     for(int i = 0,ret = 0; str[i];) {

11         if(str[i] == '~') flag = false;

12         if(isdigit(str[i])) {

13             while(isdigit(str[i])) ret = ret*10 + (str[i++]-'0');

14             s[idx].push_back(make_pair(ret-1,flag));

15             ret = 0;

16             flag = true;

17         } else i++;

18     }

19 }

20 bool check(int o){

21     for(int i = 0; i < m; ++i){

22         bool flag = false;

23         for(int j = s[i].size()-1; j >= 0; --j){

24             if(s[i][j].second == ((o>>s[i][j].first)&1)){

25                 flag = true;

26                 break;

27             }

28         }

29         if(!flag) return false;

30     }

31     return true;

32 }

33 int main(){

34     int T;

35     scanf("%d",&T);

36     while(T--){

37         scanf("%d %d",&n,&m);

38         getchar();

39         for(int i = 0; i < m; ++i){

40             s[i].clear();

41             gets(str);

42             exp2num(i);

43         }

44         bool ok = false;

45         for(int i = 0; i < (1<<n) && !ok; ++i)

46             ok = check(i);

47         puts(ok?"satisfiable":"unsatisfiable");

48     }

49     return 0;

50 }

View Code