POJ3678 Katu Puzzle

Katu Puzzle

Katu Puzzle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12601		Accepted: 4627

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".

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.

Source

POJ Founder Monthly Contest – 2008.07.27, Dagger

题解

2-SAT经典题，连边如下:

A AND B = 1	!A->A	!B->B
A AND B = 0	A->!B	B->!A
A OR B = 1	!A->B	!B->A
A OR B = 0	A->!A	B->!B
A XOR B = 1	A->!B	!B->A	!A->B	B->!A
A XOR B = 0	A->B	B->A	!A->!B	!B->!A

然后tarjan求SCC判断有无解即可。

时间复杂度\(O(n+m)\)

#include<iostream>
#include<vector>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
    rg T data=0,w=1;rg char ch=getchar();
    for(;!isdigit(ch);ch=getchar())if(ch=='-') w=-w;
    for(;isdigit(ch);ch=getchar()) data=data*10+ch-'0';
    return data*w;
}
template<class T>il T read(rg T&x) {return x=read<T>();}
typedef long long ll;
using namespace std;

co int N=1006;
int n,m,dfn[N],low[N],num,st[N],top,c[N],cnt;
bool ins[N];
vector<int> e[N*2];

il void add(int x,int y){
	e[x].push_back(y);
}
void tarjan(int x){
	dfn[x]=low[x]=++num;
	st[++top]=x,ins[x]=1;
	for(unsigned i=0;i<e[x].size();++i){
		int y=e[x][i];
		if(!dfn[y]){
			tarjan(y);
			low[x]=min(low[x],low[y]);
		}
		else if(ins[y]) low[x]=min(low[x],dfn[y]);
	}
	if(dfn[x]==low[x]){
		++cnt;
		int y;
		do y=st[top--],ins[y]=0,c[y]=cnt;
		while(x!=y);
	}
}
int main(){
	read(n),read(m);
	for(int a,b,c;m--;){
		char s[6];
		read(a),read(b),read(c),scanf("%s",s);
		if(s[0]=='A'){
			if(c) add(a,a+n),add(b,b+n);
			else add(a+n,b),add(b+n,a);
		}
		else if(s[0]=='O'){
			if(c) add(a,b+n),add(b,a+n);
			else add(a+n,a),add(b+n,b);
		}
		else{
			if(c) add(a,b+n),add(b,a+n),add(a+n,b),add(b+n,a);
			else add(a,b),add(b,a),add(a+n,b+n),add(b+n,a+n);
		}
	}
	for(int i=0;i<n;++i)
		if(!dfn[i]) tarjan(i);
	for(int i=0;i<n;++i)
		if(c[i]==c[i+n]) return puts("NO"),0;
	return puts("YES"),0;
}