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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> CodeForces 546C(隊列),codeforces546c

CodeForces 546C(隊列),codeforces546c

編輯:C++入門知識

CodeForces 546C(隊列),codeforces546c


   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

     

題解:
兩個士兵打牌,按順序出牌,若a的牌比b大,則把b的牌放到a的最後,再把a的這張牌放到b的那張排後面。很明顯此題是隊列問題可設置兩個隊列,按照題目意思模擬一下即可,注意在循環一定次數下仍然沒有結果,則無解;
另外就是可以直接標記狀態搜索,可只標記第一個,也可以標記全部,但是自己寫的有點亂,所以只介紹第一種了
隊列用法:

#include<iostream>
#include<queue>
using namespace std;
const int N=1000;
queue <int>x,y;
int main()
{
  int i,t,a,b,n,m;
  cin>>t;
  cin>>n;
for( i=0; i<n; i++)
{
  cin>>a;
  x.push(a);
}
  cin>>m;
for( i=0; i<m; i++)
{
  cin>>b;
  y.push(b);
}
  int s=0;
while(!x.empty()&&!y.empty())
{
  s++;
  if(s>N) break;
  int p=x.front();
  int q=y.front();
  x.pop();
  y.pop();
if(p<q)
{
  y.push(p);
  y.push(q);
}
if(p>q)
{
  x.push(q);
  x.push(p);
}
}
if(x.empty())
  cout<<s<<" "<<"2"<<endl;
else if(y.empty())
  cout<<s<<" "<<"1"<<endl;
else cout<<"-1"<<endl;


return 0;
}

 

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