Puzzle floor PMUs - 3678 (2 - hour)

With N variables

Given the M equations, each equation of the form

Ask whether there is a legal assignment to each variable, so that all equations are true.

Input Format

The first line contains two integers N and M.

Next M rows, each row containing three integers ABC, and a bit operation (AND, OR, XOR one).

Output Format

The output, if present, output "YES", otherwise a "NO"

data range

Sample input:

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

Sample output:

YES

Ideas: The meaning of the questions, the problem is a 2-SAT problem, the conversion operator, can be determined

For A [x], may be 'implemented <, x, NOT A [x], can even edge <x, x through the communicating edge X>' implemented> for NOT (A [x] AND A [y]) You need to connect two edges <x, y '> and <y, x'> be achieved, for A [x] OR A [y] need to connect two edges <x ', y> and <y', x> to achieve

So for and, or, xor three operations are:

and operations:

a and b = 0: If there is a = 1, then there must satisfy b = 0; if b = 1, then there must satisfy a = 0, namely: <a, 1, b, 0>, <b, 1, a, 0>
A = B and when. 1: a = 1 and b = 1, namely: <A, 0, A,. 1>, <B, 0, B,. 1>
or operation:

a or b = 0 when: a = 0 and b = 0, ie: <A,. 1, A, 0>, <B,. 1, B, 0>
A or B =. 1: If there is a = 0, then there must satisfies b = 1; if b = 0, then there must satisfy a = 0, namely: <A, 0, B,. 1>, <B, 0, A,. 1>
XOR operation:

a xor b = 0, the following four cases:
if a = 0, then there must satisfy b = 0, ie: <a, 0, b, 0>
if b = 0, then there must satisfy a = 0, namely: <b, 0, a, 0 >
if a = 1, then there must satisfy b = 1, namely: <a, 1, b, 1>
if b = 1, then there must satisfy a = 1, namely: <b, 1 , a,. 1>
a = B XOR. 1, the following four cases:
if a = 0, then there must satisfy b = 1, namely: <a, 0, b, 1>
if b = 0, then a must be satisfied = 1, namely: <b, 0, a, 1>
if a = 1, then there must satisfy b = 0, ie: <a, 1, b, 0>
if b = 1, then there must satisfy a = 0, i.e., : <b, 1, a, 0>

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>usingnamespace std;
#define debug(x) cout << "fuck bug " << x << "\n";
#define ios ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
constint maxn = 4e5 + 7;
typedef longlong ll;int n,m;struct edge
{int to,nxt;
}e[maxn];int head[maxn],tot;
void add(int u ,
    and [int

   




    

 v){++tot].to = v;
    e[tot].nxt = head[u];
    head[u] = tot;
}

int dfn[maxn],low[maxn],num,inStack[maxn];
int stack[maxn],top,cnt,C[maxn];
void tarjan(int x) {
    dfn[x] = low[x] = ++num;
    stack[++top] = x; inStack[x] = true;
    for (int i = head[x]; i; i = e[i].nxt) {
        int y = e[i].to;
        if (dfn[y] == 0) {
            tarjan(y);
            low[x] = min(low[x], low[y]);
        } else if (inStack[y]) {
            low[x] = min(low[x], low[y]);
        }
    }
    if (low[x] == dfn[x]) {
        cnt++;
        int z;
        do {
            z = stack[top--];
            inStack[z] = false;
            C[z] = cnt;
        } while (z != x);
    }
}

int main(int argc, char const *argv[])
{
    //cin >> n >> m;
    scanf("%d %d",&n,&m);
    for(int i = 1;i <= m ; i ++){
        int a , b , c ; char str[10];
        //cin >> a >> b >> c >> str;
        scanf("%d %d %d %s",&a,&b,&c,str);
        if(str[0] == 'A'){
            if(c == 1){
                add(a,a+n);
                add(b,b+n);
            }else{
                add(a+n,b);
                add(b+n,a);
            }
        }else if(str[0] == 'O'){
            if(c == 1){
                add(a,b+n);
                add(b,a+n);
            }else{
                add(a+n,a);
                add(b+n,b);
            }
        }else if(str[0] == 'X'){
            if(c == 1){
                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 + n; i++){
        if(!dfn[i]) tarjan(i); 
    }

    bool flag=1;
    for(int i = 0;i < n ;i ++){
        if(C[i] == C[i + n]) {
            flag = 0;
            break;
        }
    }

    if(flag) puts("YES");
    else puts("NO");

    return 0;
}