【POJ - 3678】Katu Puzzle（2-SAT之判断有解）

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.

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.

**思路：**判断是否有解问题，按照要求加边就行了。
代码：

#include
#include
#include
#include
#include
#define maxn 2005
#define maxx 5000005
#define INF 0x3f3f3f3f
using namespace std;
int n,_n,m;
int head[maxn],to[maxx],_next[maxx],edge;
inline void addEdge(int x,int y)
{
    to[++edge]=y,_next[edge]=head[x],head[x]=edge;
}
int dfn[maxn],low[maxn],_index=0;
int st[maxn],cnt=0;
bool inS[maxn];
int be[maxn],tot=0;
void tarjan(int u)
{
    dfn[u]=low[u]=++_index;
    st[++cnt]=u;inS[u]=true;
    for(int i=head[u];i;i=_next[i])
    {
        int v=to[i];
        if(!dfn[v])
        {
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        else if(inS[v])low[u]=min(low[u],dfn[v]);
    }
    if(dfn[u]==low[u])
    {
        ++tot;int now;
        do
        {
            now=st[cnt--];
            inS[now]=false;
            be[now]=tot;
        }while(u!=now);
    }
}
int main()
{
    cin>>n>>m;
    char s[10];
    int x,y,z;
    for(int i=0;i