ICM Technex 2017 and Codeforces Round #400 (Div. 1 + Div. 2, combined) D

Posted 2020-09-29 樱花落舞

Moriarty has trapped n people in n distinct rooms in a hotel. Some rooms are locked, others are unlocked. But, there is a condition that the people in the hotel can only escape when all the doors are unlocked at the same time. There are m switches. Each switch control doors of some rooms, but each door is controlled by exactly two switches.

You are given the initial configuration of the doors. Toggling any switch, that is, turning it ON when it is OFF, or turning it OFF when it is ON, toggles the condition of the doors that this switch controls. Say, we toggled switch 1, which was connected to room 1, 2 and 3 which were respectively locked, unlocked and unlocked. Then, after toggling the switch, they become unlocked, locked and locked.

You need to tell Sherlock, if there exists a way to unlock all doors at the same time.

Input

First line of input contains two integers n and m (2?≤?n?≤?105, 2?≤?m?≤?105) — the number of rooms and the number of switches.

Next line contains n space-separated integers r1,?r2,?...,?rn (0?≤?ri?≤?1) which tell the status of room doors. The i-th room is locked if ri?=?0, otherwise it is unlocked.

The i-th of next m lines contains an integer xi (0?≤?xi?≤?n) followed by xi distinct integers separated by space, denoting the number of rooms controlled by the i-th switch followed by the room numbers that this switch controls. It is guaranteed that the room numbers are in the range from 1to n. It is guaranteed that each door is controlled by exactly two switches.

Output

Output "YES" without quotes, if it is possible to open all doors at the same time, otherwise output "NO" without quotes.

Examples

input

3 3
1 0 1
2 1 3
2 1 2
2 2 3

output

NO

input

3 3
1 0 1
3 1 2 3
1 2
2 1 3

output

YES

input

3 3
1 0 1
3 1 2 3
2 1 2
1 3

output

NO

Note

In the second example input, the initial statuses of the doors are [1,?0,?1] (0 means locked, 1 — unlocked).

After toggling switch 3, we get [0,?0,?0] that means all doors are locked.

Then, after toggling switch 1, we get [1,?1,?1] that means all doors are unlocked.

It can be seen that for the first and for the third example inputs it is not possible to make all doors unlocked.

题意：一个门受到两个开关控制，问最后能不能把门全部打开（详情看输入格式和解释）

解法：

1 如果门是开的，那么控制的两个开关要么一起关，要么一起开

2 如果门是关的，那么控制的两个开关只需要开一个

3 不能全部开的条件，很幸运，经过分析发现 有一个控制关a门，也控制开a门，那么这门是打不开了（矛盾了）

4 我们把开关看做点，门看做线，没有+m看作是开门，+m看作关门，用并查集处理，找到一个开门和关门处于同一个集合的情况，矛盾不存在可行结果

#include<bits/stdc++.h>
typedef long long LL;
typedef unsigned long long ULL;
using namespace std;
const int maxn=1e5;
vector<int>Ve[2*maxn];
int tree[2*maxn];
int Find(int x)
{
    if(x==tree[x])
        return x;
    return tree[x]=Find(tree[x]);
}

void Merge(int x,int y)
{
    int fx=Find(x);
    int fy=Find(y);
    if(fx!=fy)
        tree[fx]=fy;
}
int door[maxn];
int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=2*m;i++){
        tree[i]=i;
    }
    for(int i=1;i<=n;i++){
        scanf("%d",&door[i]);
    }
    for(int i=1;i<=m;i++){
        int num;
        scanf("%d",&num);
        for(int j=1;j<=num;j++){
            int x;
            scanf("%d",&x);
            Ve[x].push_back(i);
        }
    }
    for(int i=1;i<=n;i++){
        int x=Ve[i][0];
        int y=Ve[i][1];
        if(door[i]){
            Merge(x,y);
            Merge(x+m,y+m);
        }else{
            Merge(x,y+m);
            Merge(x+m,y);
        }
    }
    int flag=1;
    for(int i=1;i<=m;++i){
        if(Find(i)==Find(i+m)){
            flag=0;
            break;
        }
    }
    if(flag)
        printf("YES\n");
    else
        printf("NO\n");
    return 0;
}