Algorithm: POJ 3678 katu Puzzle (2-sat judgment)

"The main effect of the topic"

There is a direction graph G (v,e), each side E (A,B) has a bitwise operator OP (and, or or XOR) and a value C (0 or 1).

Ask if you can assign a value of x (0 or 1) to each point on the graph so that each edge e (a,b) satisfies Xa op Xb = c

Ideas

Each point can take only 0 or 1, obviously the 2-SAT model.

The key is how to build the edge.

For two points A and B, a has two values a1=0,a2=1, B also has two values b1=0, B2=1.

Then enumerate all the relationships of these two points

(A1, B1)

(A1, B2)

(A2, B1)

(A2, B2)

Then according to the bitwise operator to see if each relationship is eligible or not, if not meet the relationship when the contradiction is, to add two edges

The assumption is the relationship (A1,B1) Contradiction, then we should add edge a1->b2, B1->A2 can be.

by analogy.

Code

#include <iostream> #include <cstdio> #include <cstring> using namespace std;  
    
typedef long long Int64;  
const int MAXN = 2010;  
const int VN = MAXN*2;  
    
const int EN = 4000010;  
    
int n, m;  
        Class graph{Public:void init () {size = 0;  
    Memset (Head,-1, sizeof (head));  
        } void Addedge (int u, int v) {e[size].v = v;  
        E[size].next = Head[u];  
    Head[u] = size++;  
    } public:int size;  
    int HEAD[VN];   
    struct edge{int V, next;  
}e[en];  
    
}g;   
        Class two_sat{Public:bool Check (const graph&g, const int n) {SCC (g, N);  
        for (int i=0; i<n; ++i) if (belong[i*2] = = Belong[i*2+1]) return false;  
    return true;  
        } private:void Tarjan (const graph&g, const int u) {int V;  
        Dfn[u] = low[u] = ++idx;  
        sta[top++] = u; Instack[u] = true;  
            for (int e=g.head[u]; E!=-1 e=g.e[e].next) {v = g.e[e].v;  
                if (dfn[v] = =-1) {Tarjan (g, v);  
            Low[u] = min (Low[u], low[v]);  
            }else if (Instack[v]) {Low[u] = min (Low[u], dfn[v]);  
            } if (low[u] = = Dfn[u]) {++bcnt;  
                do{v = sta[--top];  
                INSTACK[V] = false;  
            BELONG[V] = bcnt;  
        }while (U!= v);    
        } void SCC (const GRAPH&G, const int n) {top = idx = bcnt = 0;  
        memset (DFN,-1, sizeof (DFN));  
        memset (instack, 0, sizeof (instack));  
        for (int i=0; i<2*n; ++i) {if (dfn[i] = = 1) Tarjan (g, I);  
    } private:int top, idx, bcnt;  
    int STA[VN];  
    int DFN[VN];  
    int LOW[VN];  
    int BELONG[VN];  
BOOL INSTACK[VN];}sat;  
    int main () {int u, V, W;  
    Char op[5];  
        while (~SCANF ("%d%d", &n, &m)) {g.init ();       
            for (int i=0; i<m; ++i) {scanf ("%d%d%d%s", &u, &v, &w, op);     if (!strcmp (OP, "and")) {if (W) {G.addedge (u*2, v*2+1), G.addedge (v*2, u*2+1);   0, 0 G.addedge (u*2, v*2), G.addedge (v*2+1, u*2+1);       0, 1 G.addedge (u*2+1, v*2+1), G.addedge (v*2, u*2);     1, 0}else{G.addedge (u*2+1, v*2), G.addedge (v*2+1, u*2); 1, 1}}else if (!strcmp (OP, "OR") {if (W) {G.A     Ddedge (u*2, v*2+1), G.addedge (v*2, u*2+1);   0, 0}else{g.addedge (u*2, v*2), G.addedge (v*2+1, u*2+1); 0, 1 G.addedge (u*2+1, v*2+1), G.addedge (v*2, u*2);     1, 0 G.addedge (u*2+1, v*2), G.addedge (v*2+1, u*2); 1, 1}}else{//XOR if (w) {G.addedge (u*2, V     *2+1), G.addedge (v*2, u*2+1);     0, 0 G.addedge (u*2+1, v*2), G.addedge (v*2+1, u*2);   1, 1}else{g.addedge (u*2, v*2), G.addedge (v*2+1, u*2+1);       0, 1 G.addedge (u*2+1, v*2+1), G.addedge (v*2, u*2);  
        1, 0}}} if (Sat.check (g, N)) puts ("YES");  
    Else puts ("NO");  
return 0; }

See more highlights of this column: http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/