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 = 200;
char str[maxn];
vector< pii >s[maxn];
int n,m;
void exp2num(int idx) {
    bool flag = true;
    for(int i = 0,ret = 0; str[i];) {
        if(str[i] == '~') flag = false;
        if(isdigit(str[i])) {
            while(isdigit(str[i])) ret = ret*10 + (str[i++]-'0');
            s[idx].push_back(make_pair(ret-1,flag));
            ret = 0;
            flag = true;
        } else i++;
    }
}
bool check(int o){
    for(int i = 0; i < m; ++i){
        bool flag = false;
        for(int j = s[i].size()-1; j >= 0; --j){
            if(s[i][j].second == ((o>>s[i][j].first)&1)){
                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 = 0; i < m; ++i){
            s[i].clear();
            gets(str);
            exp2num(i);
        }
        bool ok = false;
        for(int i = 0; i < (1<<n) && !ok; ++i)
            ok = check(i);
        puts(ok?"satisfiable":"unsatisfiable");
    }
    return 0;
}