### Problem

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

### Analysis

2-SAT模板题，根据输入的类型加入条件即可。注意addClause(int x, int xval, int y, int yval)为“x为xval或者y为yval”。

### Code

```#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

const int maxn = 1024;

struct TwoSAT {
int n, c, S[maxn*2];
vector<int> G[maxn*2];
bool mark[maxn*2];

void init(int n) {
this->n = n;
for(int i = 0; i < n * 2; ++i)
G[i].clear();
memset(mark, 0, sizeof(mark));
}
bool dfs(int u) {
if(mark[u ^ 1]) return false;
if(mark[u]) return true;
mark[u] = true;
S[c++] = u;
for(int i = 0; i < G[u].size(); ++i)
if(!dfs(G[u][i])) return false;
return true;
}
void addClause(int x, int xval, int y, int yval) {
x = x * 2 + xval;
y = y * 2 + yval;
G[x^1].push_back(y);
G[y^1].push_back(x);
}
bool solve() {
for(int i = 0; i < n * 2; i += 2) {
if(!mark[i] && !mark[i+1]) {
c = 0;
if(!dfs(i)) {
while(c > 0) mark[S[--c]] = false;
if(!dfs(i+1)) return false;
}
}
}
return true;
}
}sat;

int main() {
int N, M; scanf("%d%d", &N, &M);
int a, b, c; char op[10];
sat.init(N);
while(M--) {
scanf("%d%d%d%s", &a, &b, &c, op);
if(op[0] == 'A') {
if(c == 1) {
sat.addClause(a, 1, b, 1);
sat.addClause(a, 0, b, 1);
sat.addClause(a, 1, b, 0);
} else if(c == 0) {
sat.addClause(a, 0, b, 0);
}
} else if(op[0] == 'O') {
if(c == 1) {
sat.addClause(a, 1, b, 1);
} else if(c == 0) {
sat.addClause(a, 0, b, 0);
sat.addClause(a, 1, b, 0);
sat.addClause(a, 0, b, 1);
}
} else if(op[0] == 'X') {
if(c == 1) {
sat.addClause(a, 1, b, 1);
sat.addClause(a, 0, b, 0);
} else if(c == 0) {
sat.addClause(a, 1, b, 0);
sat.addClause(a, 0, b, 1);
}
}
}
puts(sat.solve() ? "YES" : "NO");
return 0;
}
```