有向图双联通性与逻辑判定 - 石头剪刀布
• 玩剪刀石头布，n把
• 给了A这n把都出了什么，问你B能否会赢
• 其中A会限制B某些局数出的要相同
• 某些局数出的要不同
• 只要B满足他的限制，并且没没有输掉任何一把就算赢

A每局有2个状态，赢和平
令xi表示第i局A是否赢
2sat分8种情况讨论

#include<bits/stdc++.h>
using namespace std;
#define
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#define
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
ll Abs(ll x){return (x<0)?-x:x;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 

#define
struct TwoSat{
    int n;
    vi G[MAXN*2]; 
    bool mark[MAXN*2];
    int S[MAXN*2],c;
    bool dfs(int x) {
        if (mark[x^1]) return 0;
        if (mark[x]) return 1;
        mark[x]=1;
        S[c++]=x;
        int sz=SI(G[x]);
        Rep(i,sz) if (!dfs(G[x][i])) return 0;
        return 1;
    }
    void init(int n) {
        this -> n = n;
        Rep(i,n*2) G[i].clear();
        MEM(mark)
    }
    //x = xval or y = yval
    void add_clause(int x,int xval,int y,int yval) {
        x=x*2+xval;
        y=y*2+yval;
        G[x^1].pb(y);
        G[y^1].pb(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) mark[S[--c]] = 0;
                    if (!dfs(i+1)) return 0;
                }
            }
        }
        return 1;
    }
}solver;
int n,m;
int a[MAXN];
// win,los
int h[4]={0,2,3,1};
int main()
{
//  freopen("C.in","r",stdin);
//  freopen(".out","w",stdout);
    int T=read();
    For(kcase,T) {
        n=read(),m=read();
        solver.init(3*n);
        For(i,n) {
            a[i]=read();
        }
        For(i,m) {
            int u=read(),v=read(),k=read();
            int wu=h[a[u]],lu=a[u],wv=h[a[v]],lv=a[v];
            if (!k) {
                if (wu!=wv) solver.add_clause(u,0,v,0); 
                if (wu!=lv) solver.add_clause(u,0,v,1); 
                if (lu!=wv) solver.add_clause(u,1,v,0); 
                if (lu!=lv) solver.add_clause(u,1,v,1); 
            } else {
                if (wu==wv) solver.add_clause(u,0,v,0); 
                if (wu==lv) solver.add_clause(u,0,v,1); 
                if (lu==wv) solver.add_clause(u,1,v,0); 
                if (lu==lv) solver.add_clause(u,1,v,1); 
            }
        }

        Pr(kcase)
        if (solver.solve()) puts("yes"); else puts("no");
    }   
    return 0;
}