poj 3678 Katu Puzzle(2-sat)

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 Vi a value Xi (0 ≤ Xi ≤ 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:

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

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 a line containing "YES" or "NO".

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

YES

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

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
#include <stack>
using namespace std;
#define PI acos(-1.0)
#define max(a,b) (a) > (b) ? (a) : (b)
#define min(a,b) (a) < (b) ? (a) : (b)
#define ll long long
#define eps 1e-10
#define MOD 1000000007
#define N 1006
#define inf 1e12
int n,m;
vector<int> e
;

int tot;
int head
;
int vis
;
int tt;
int scc;
stack<int>s;
int dfn
,low
;
int col
;
struct Node
{
int from;
int to;
int next;
}edge[N*N];
void init()
{
tot=0;
scc=0;
tt=0;
memset(head,-1,sizeof(head));
memset(dfn,-1,sizeof(dfn));
memset(low,0,sizeof(low));
memset(vis,0,sizeof(vis));
memset(col,0,sizeof(col));
}
void add(int s,int u)//邻接矩阵函数
{
edge[tot].from=s;
edge[tot].to=u;
edge[tot].next=head[s];
head[s]=tot++;
}
void tarjan(int u)//tarjan算法找出图中的所有强连通分支
{
dfn[u] = low[u]= ++tt;
vis[u]=1;
s.push(u);
int cnt=0;
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(dfn[v]==-1)
{
//    sum++;
tarjan(v);
low[u]=min(low[u],low[v]);
}
else if(vis[v]==1)
low[u]=min(low[u],dfn[v]);
}
if(dfn[u]==low[u])
{
int x;
scc++;
do{
x=s.top();
s.pop();
col[x]=scc;
vis[x]=0;
}while(x!=u);
}
}
bool two_sat(){

for(int i=0;i<2*n;i++){
if(dfn[i]==-1){
tarjan(i);
}
}
for(int i=0;i<n;i++){
if(col[2*i]==col[2*i+1]){
return false;
}
}
return true;
}
int main()
{
while(scanf("%d%d",&n,&m)==2){
init();
for(int i=0;i<N;i++) e[i].clear();
while(!s.empty()){
s.pop();
}
int a,b,c;
char s[6];
for(int i=0;i<m;i++){
scanf("%d%d%d%s",&a,&b,&c,s);
if(s[0]=='A'){
if(c==1){
//e[2*a+1].push_back(2*a);
//e[2*b+1].push_back(2*b);
add(2*a+1,2*a);
add(2*b+1,2*b);
}
else{
//e[2*a].push_back(2*b+1);
//e[2*b].push_back(2*a+1);
add(2*a,2*b+1);
add(2*b,2*a+1);
}
}
else if(s[0]=='O'){
if(c==1){
//e[2*a+1].push_back(2*b);
//e[2*b+1].push_back(2*a);
add(2*a+1,2*b);
add(2*b+1,2*a);
}
else{
//e[2*a].push_back(2*a+1);
//e[2*b].push_back(2*b+1);
add(2*a,2*a+1);
add(2*b,2*b+1);
}
}
}
if(two_sat()){
printf("YES\n");
}
else{
printf("NO\n");
}
}
return 0;
}