[POJ3678] Katu Puzzle | 2-SAT 入门

Posted 2021-12-08 PushyTao

Description

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

Xa op Xb = 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 $(1\le N\le 1000)$ and M,$(0\le M\le 1,000,000)$ indicating the number of vertices and edges.
The following M lines contain three integers a $(0\le a<N),b(0\le b<N)$, c and an operator op each, describing the edges.

Output

Output a line containing “YES” or “NO”.

Sample Input

4 4
0 1 1 AND
1 2 1 OR
3 2 0 AND
3 0 0 XOR
Sample Output

YES
Hint

X0 = 1, X1 = 1, X2 = 0, X3 = 1.

int n, m;
struct node {
	int to,nex;
} e[maxm << 2];
int cnt = 0, head[maxn << 1];
bool inStk[maxn << 1];
int low[maxn << 1], dfn[maxn << 1],dfc, pos[maxn << 1];
void init() {
	dfc = cnt = 0;
	for(int i=0; i <=2* n; i++) {
		head[i] = -1;
		low[i] = dfn[i] = 0;
	}
}
stack<int> stk;
void add(int u,int v) {
	e[cnt].to = v;
	e[cnt].nex = head[u];
	head[u] = cnt ++;
}
int cntSCC;
void Tarjan(int u) {
//	cout << u << endl;
	dfn[u] = low[u] = ++ dfc;
	inStk[u] = 1;
	stk.push(u);
	for(int i=head[u]; ~i; i=e[i].nex) {
		itn to = e[i].to;
		if(!dfn[to]) {
			Tarjan(to);
			low[u] = min(low[u],low[to]);
		} else if(inStk[to] && !pos[to]) {
			low[u] = min(low[u],dfn[to]);
		}
	}
	if(low[u] == dfn[u]) {
		int tp;
		++ cntSCC;
		do {
			tp = stk.top();
			stk.pop();
			inStk[tp] = 0;
			pos[tp] = cntSCC;
		} while(tp != u);
	}
}
int a,b,c;
string op;
int main() {
	n=read,m=read;
	init();
	for(int i=1; i<=m; i++) {
		a = read,b = read,c = read;
		cin >> op;
		int u=a * 2,u1=2 * a+1;
		int v=b * 2,v1=2 * b+1;

		if(op == "AND") {
			if(c == 1) {/// u,v
				add(u1,u);
				add(v1,v);
			} else {
//				add(u,u1);
				add(u,v1);
//				add(v,u1);
				add(v,u1);
			}
		} else if(op == "OR") {
			if(c == 1) {
//				add(u1,u);
				add(u1,v);
//				add(v1,v);
				add(v1,u);
			} else {
				add(u,u1);
				add(v,v1);
			}
		} else if(op == "XOR") {
			if(c == 1) {
				add(u,v1);
				add(v,u1);
				add(u1,v);
				add(v1,u);
			} else {
				add(u,v);
				add(v,u);
				add(u1,v1);
				add(v1,u1);
			}
		}
	}
	for(int i=0; i <= 2*n; i++) {
		if(!dfn[i]) Tarjan(i);
	}
	for(int i=0; i <= 2*n; i++) {
		if(pos[i] == pos[i+1]) {
			puts("NO");
			return 0;
		}
		i ++;
	}
	puts("YES");
	return 0;
}