CodeForces 546C（队列）

　　　CodeForces 546C Soldier and Cards

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

 

Description

Two bored soldiers are playing card war. Their card deck consists of exactly n cards, numbered from 1 to n, all values are different. They divide cards between them in some manner, it's possible that they have different number of cards. Then they play a "war"-like card game.

The rules are following. On each turn a fight happens. Each of them picks card from the top of his stack and puts on the table. The one whose card value is bigger wins this fight and takes both cards from the table to the bottom of his stack. More precisely, he first takes his opponent's card and puts to the bottom of his stack, and then he puts his card to the bottom of his stack. If after some turn one of the player's stack becomes empty, he loses and the other one wins.

You have to calculate how many fights will happen and who will win the game, or state that game won't end.

Input

First line contains a single integer n (2 ≤ n ≤ 10), the number of cards.

Second line contains integer k₁ (1 ≤ k₁ ≤ n - 1), the number of the first soldier's cards. Then follow k₁ integers that are the values on the first soldier's cards, from top to bottom of his stack.

Third line contains integer k₂ (k₁ + k₂ = n), the number of the second soldier's cards. Then follow k₂ integers that are the values on the second soldier's cards, from top to bottom of his stack.

All card values are different.

Output

If somebody wins in this game, print 2 integers where the first one stands for the number of fights before end of game and the second one is 1 or 2 showing which player has won.

If the game won't end and will continue forever output  - 1.

Sample Input

Input

4
2 1 3
2 4 2

Output

6 2

Input

3
1 2
2 1 3

Output

-1

题解：两个士兵打牌，按顺序出牌，若a的牌比b大，则把b的牌放到a的最后，再把a的这张牌放到b的那张排后面。很明显此题是队列问题可设置两个队列，按照题目意思模拟一下即可，注意在循环一定次数下仍然没有结果，则无解；另外就是可以直接标记状态搜索，可只标记第一个，也可以标记全部，但是自己写的有点乱，所以只介绍第一种了

#include<iostream>
#include<queue>
using namespace std;
const int N=;
queue <int>x,y;
int main()
{
    int i,t,a,b,n,m;
    cin>>t;
    cin>>n;
    for( i=; i<n; i++)
    {
        cin>>a;
        x.push(a);
    }
    cin>>m;
    for( i=; i<m; i++)
    {
        cin>>b;
        y.push(b);
    }
    int s=;
    while(!x.empty()&&!y.empty())
    {
        s++;
        if(s>N) break;
        int q=x.front();
        int p=y.front();
        x.pop();
        y.pop();
        if(q<p)
        {
            y.push(q);
            y.push(p);
        }
        if(q>p)
        {
            x.push(p);
            x.push(q);
        }
    }
        if(x.empty())
           cout<<s<<" "<<""<<endl;
        else if(y.empty())
           cout<<s<<" "<<""<<endl;
        else cout<<"-1"<<endl;

    return ;
}