POJ 3678 Katu Puzzle（2-SAT）

Description

Katu Puzzle is presented as a directed graph G(V, E) with each edge e(a, b) labeled by a boolean operator op (one of AND, OR, XOR) and an integer c (0 ≤ c ≤ 1). One Katu is solvable if one can find each vertex V_i a value X_i (0 ≤ X_i ≤ 1) such that for each edge e(a, b) labeled by op and c, the following formula holds:

 X_a op X_b = c

The calculating rules are:

AND 0 1 0 0 0 1 0 1

OR 0 1 0 0 1 1 1 1

XOR 0 1 0 0 1 1 1 0

Given a Katu Puzzle, your task is to determine whether it is solvable.

Input

The first line contains two integers N (1 ≤ N ≤ 1000) and M,(0 ≤ M ≤ 1,000,000) indicating the number of vertices and edges.
The following M lines contain three integers a (0 ≤ a < N), b(0 ≤ b < N), c and an operator op each, describing the edges.

Output

Output a line containing "YES" or "NO"

 

题目大意：每条边给出一个OP和一个ANS，要求每条边的连点A、B，符合A OP B == ANS，其中A、B、ANS都为0或1

思路：2-SAT问题

A AND B == 0则A真→B假，B真→A假

A AND B == 1则A、B都为真，即B假→B真，A假→A真（即不能出现A假推出A真的情况，A假不能被选上，在强连通图中，若A真通过多步推出A假，即成环，无解）

A OR B == 0则A、B都为假，即B真→B假，A真→A假（同上）

A OR B == 1则A假→B真，B假→A真

A XOR B == 0则A真↔B真，A假↔B假

A XOR B == 1则A真↔B假，A假↔B真

 

 1 #include <cstdio>

 2 #include <cstring>

 3 using namespace std;

 4 

 5 const int MAXN = 2010;

 6 const int MAXM = 1000010*4;

 7 

 8 struct TwoSAT{

 9     int n, ecnt, dfs_clock, scc_cnt;

10     int St[MAXN], c;

11     int head[MAXN], sccno[MAXN], lowlink[MAXN], pre[MAXN];

12     int next[MAXM], to[MAXM];

13 

14     void init(int nn){

15         n = nn;

16         ecnt = 2; dfs_clock = scc_cnt = 0;

17         memset(head,0,sizeof(head));

18     }

19 

20     void addEdge(int x, int y){

21         to[ecnt] = y; next[ecnt] = head[x]; head[x] = ecnt++;

22     }

23 

24     void addEdge2(int x, int y){

25         addEdge(x,y); addEdge(y,x);

26     }

27 

28     void dfs(int u){

29         lowlink[u] = pre[u] = ++dfs_clock;

30         St[++c] = u;

31         for(int p = head[u]; p; p = next[p]){

32             int &v = to[p];

33             if(!pre[v]){

34                 dfs(v);

35                 if(lowlink[u] > lowlink[v]) lowlink[u] = lowlink[v];

36             }else if(!sccno[v]){

37                 if(lowlink[u] > pre[v]) lowlink[u] = pre[v];

38             }

39         }

40         if(lowlink[u] == pre[u]){

41             ++scc_cnt;

42             while(true){

43                 int x = St[c--];

44                 sccno[x] = scc_cnt;

45                 if(x == u) break;

46             }

47         }

48     }

49 

50     bool solve(){

51         for(int i = 0; i < n; ++i)

52             if(!pre[i]) dfs(i);

53         for(int i = 0; i < n; i += 2)

54             if(sccno[i] == sccno[i^1]) return false;

55         return true;

56     }

57 

58 } G;

59 

60 int main(){

61     int n, m, a, b, c;

62     char d[5];

63     while(scanf("%d%d",&n,&m)!=EOF){

64         G.init(2*n);

65         while(m--){

66             scanf("%d%d%d%s",&a,&b,&c,d);

67             if(d[0] == 'A' && c == 0){//a&b=false

68                 G.addEdge(a*2, b*2+1);

69                 G.addEdge(b*2, a*2+1);

70             }

71             if(d[0] == 'A' && c == 1){//a&b=true

72                 G.addEdge(a*2+1, a*2);

73                 G.addEdge(b*2+1, b*2);

74             }

75             if(d[0] == 'O' && c == 0){//a|b=false

76                 G.addEdge(a*2, a*2+1);

77                 G.addEdge(b*2, b*2+1);

78             }

79             if(d[0] == 'O' && c == 1){//a|b=true

80                 G.addEdge(b*2+1, a*2);

81                 G.addEdge(a*2+1, b*2);

82             }

83             if(d[0] == 'X' && c == 0){//a^b=false

84                 G.addEdge2(a*2, b*2);

85                 G.addEdge2(a*2+1, b*2+1);

86             }

87             if(d[0] == 'X' && c == 1){//a^b=true

88                 G.addEdge2(a*2, b*2+1);

89                 G.addEdge2(a*2+1, b*2);

90             }

91         }

92         if(G.solve()) printf("YES\n");

93         else printf("NO\n");

94     }

95     return 0;

96 }

141MS