您需要输出一行结果，你将采取最优的杀敌顺序，当你胜利的时候，此时还可以抵挡 t 次攻击，这时候需要您输出"Win! I still have t."， t为一个整数。或者，你被消灭了，这时候你需要输出 "Game over!"。(输出不含引号)。 Win! I still have t. Win! I still have t. Win! I still have t. 因为这个点卡我十分钟可恶！！！

本题拿BFS+优先队列提前击杀最近的即可解决

ll t[505][505];
ll ans[505][505];
ll vis[505][505];
ll dx[4]={0,0,1,-1};
ll dy[4]={1,-1,0,0};
void solve()
{
	int q,n,x0,y0,didang; cin>>q>>n>>x0>>y0>>didang;
    vec<pll>a(q);
	Fo(i,0,q)
    {
        int x,y; cin>>x>>y;
        a[i]=make_pair(x,y);
    }
	For(i,1,n)
        For(j,1,n)
            cin>>t[i][j];
    queue<pll>dl;
    ans[x0][y0]=0;
    vis[x0][y0]=1;
    dl.p(make_pair(x0,y0));
    while(!dl.empty())
    {
        ll dqx,dqy;
        dqx=dl.front().first;dqy=dl.front().second;
        dl.pop();
        Fo(i,0,4)
        {
            if((dqx+dx[i]>=1&&dqx+dx[i]<=n)&&(dqy+dy[i]>=1&&dqy+dy[i]<=n)&&vis[dqx+dx[i]][dqy+dy[i]]==0&&t[dqx+dx[i]][dqy+dy[i]]==0)
            {
                dl.p(make_pair(dqx+dx[i],dqy+dy[i]));
                ans[dqx+dx[i]][dqy+dy[i]]=ans[dqx][dqy]+1;
                vis[dqx+dx[i]][dqy+dy[i]]=1;
            }
        }
    }
    priority_queue<ll,vec<ll>,greater<>>pq;
    Fo(i,0,q)
    {
        pq.p(ans[a[i].first][a[i].second]);
        //cout<<a[i].first<<' '<<a[i].second<<' '<<ans[a[i].first][a[i].second]<<'\n';
    }
    int qj=0;
    Fo(i,0,q)
    {
        if(pq.top()-qj>0)
        {
            pq.pop();
            qj++;
        }
        else if(didang>=0)
        {   
            didang-=((pq.top()-qj)+1);
            pq.pop();
            qj++;
        }
        else 
        {
            cout<<"Game over!";
            return;
        }
    }
    if(didang>=0)
        cout<<"Win! I still have "<<didang<<'.';
    else
        cout<<"Game over!";
    return;
    
}
int main()
{
    AC;
	int t=1;
	//cin>>t;
	while(t--){solve();}
    return 0;
}