2-SAT模板（修改自LRJ的模板） 

  1 /**
  2 *2-SAT模板，Modified Edition of LRJ
  3 *输入：按照法则添加边(参数为2*i或者2*i+1)
  4 *运行：先init(n),再add(),再solve()
  5 *输出：mark[](1表示选中),solve()(是否有解)
  6 */
  7 const int maxn = 0;
  8 struct TwoSAT
  9 {
 10     int n;
 11     vector<int> G[maxn*2];
 12     bool mark[maxn*2];
 13     int S[maxn*2], c;
 14 
 15     bool dfs(int x)
 16     {
 17         if (mark[x^1]) return false;
 18         if (mark[x]) return true;
 19         mark[x] = true;
 20         S[c++] = x;
 21         for (int i = 0; i < G[x].size(); i++)
 22             if (!dfs(G[x][i])) return false;
 23         return true;
 24     }
 25 
 26     void init(int n)
 27     {
 28         this->n = n;
 29         for (int i = 0; i < n*2; i++) G[i].clear();
 30         memset(mark, 0, sizeof(mark));
 31     }
 32 
 33     /// x AND y = 1
 34     void add_and_one(int x,int y)
 35     {
 36         G[x^1].push_back(y);
 37         G[y^1].push_back(x);
 38         G[x].push_back(y);
 39         G[y^1].push_back(x^1);
 40         G[y].push_back(x);
 41         G[x^1].push_back(y^1);
 42     }
 43 
 44     /// x AND y = 0
 45     void add_and_zero(int x,int y)
 46     {
 47         G[x].push_back(y^1);
 48         G[y].push_back(x^1);
 49     }
 50 
 51     /// x OR y = 1
 52     void add_or_one(int x,int y)
 53     {
 54         G[x^1].push_back(y);
 55         G[y^1].push_back(x);
 56     }
 57 
 58     /// x OR y = 0
 59     void add_or_zero(int x,int y)
 60     {
 61         G[x].push_back(y^1);
 62         G[y].push_back(x^1);
 63         G[x].push_back(y);
 64         G[y^1].push_back(x^1);
 65         G[x^1].push_back(y^1);
 66         G[y].push_back(x);
 67     }
 68 
 69     /// x XOR y = 1
 70     void add_xor_one(int x,int y)
 71     {
 72         G[x^1].push_back(y);
 73         G[y^1].push_back(x);
 74         G[x].push_back(y^1);
 75         G[y].push_back(x^1);
 76     }
 77 
 78     /// x XOR y = 0
 79     void add_xor_zero(int x,int y)
 80     {
 81         G[x^1].push_back(y^1);
 82         G[y].push_back(x);
 83         G[x].push_back(y);
 84         G[y^1].push_back(x^1);
 85     }
 86 
 87     /// x -> y
 88     void add_to(int x,int y)
 89     {
 90         G[x].push_back(y);
 91         G[y^1].push_back(x^1);
 92     }
 93 
 94     bool solve()
 95     {
 96         for(int i = 0; i < n*2; i += 2)
 97             if(!mark[i] && !mark[i+1])
 98             {
 99                 c = 0;
100                 if(!dfs(i))
101                 {
102                     while(c > 0) mark[S[--c]] = false;
103                     if(!dfs(i+1)) return false;
104                 }
105             }
106         return true;
107     }
108 };
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posted @ 2013-08-14 16:55  xysmlx  阅读(206)  评论(0)  编辑  收藏  举报