Katu Puzzle

Time Limit: 1000MS

 

Memory Limit: 65536KB

 

64bit IO Format: %I64d & %I64u

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Status

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
Vi a value Xi (0 ≤ Xi ≤ 1) such that for each edge
e(a, b) labeled by op and c, the following formula holds:

XaopXb = 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

_X_0 = 1, _X_1 = 1,
_X_2 = 0, _X_3 = 1.

#include
#include
#include
#include
#include
#include
using namespace std;
#define MAX 1000000+10
int low[MAX],dfn[MAX];
int sccno[MAX],m,n;
int scc_cnt,dfs_clock;
bool Instack[MAX];
vectorG[MAX];
stacks;
void init()
{
for(int i=0;i<2*n;i++)
G[i].clear();
}
void getmap()
{
while(m--)
{
int a,b,c;
char op[5];
memset(op,'\0',sizeof(op));
scanf("%d%d%d%s",&a,&b,&c,op);
if(op[0]=='A')
{
if(c==1)
{
G[a+n].push_back(a);
G[b+n].push_back(b);
}
else
{
G[a].push_back(b+n);
G[b].push_back(a+n);
}
}
else if(op[0]=='O')
{
if(c==1)
{
G[a+n].push_back(b);
G[b+n].push_back(a);
}
else
{
G[a].push_back(a+n);
G[b].push_back(b+n);
}
}
else if(op[0]=='X')
{
if(c==1)
{
G[a+n].push_back(b);
G[b+n].push_back(a);
G[a].push_back(b+n);
G[b].push_back(a+n);
}
else
{
G[a].push_back(b);
G[b].push_back(a);
G[a+n].push_back(b+n);
G[b+n].push_back(a+n);
}
}
}
}
void tarjan(int u,int fa)
{
int v;
low[u]=dfn[u]=++dfs_clock;
Instack[u]=true;
s.push(u);
for(int i=0;i<G[u].size();i++)
{
v=G[u][i];
if(!dfn[v])
{
tarjan(v,u);
low[u]=min(low[v],low[u]);
}
else if(Instack[v])
low[u]=min(low[u],dfn[v]);
}
if(low[u]==dfn[u])
{
++scc_cnt;
for(;;)
{
v=s.top();
s.pop();
Instack[v]=false;
sccno[v]=scc_cnt;
if(v==u) break;
}
}
}
void find(int l,int r)
{
memset(low,0,sizeof(low));
memset(dfn,0,sizeof(dfn));
memset(sccno,0,sizeof(sccno));
memset(Instack,false,sizeof(Instack));
scc_cnt=dfs_clock=0;
for(int i=l;i<=r;i++)
if(!dfn[i]) tarjan(i,-1);
}
void solve()
{
for(int i=0;i<n;i++)
{
if(sccno[i]==sccno[i+n])
{
printf("NO\n");
return ;
}
}
printf("YES\n");
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
init();
getmap();
find(0,2*n-1);
solve();
}
return 0;
}