Traffic hdu-6573

Avin is observing the cars at a crossroads. He finds that there are n cars running in the east-west direction with the i-th car passing the intersection at time ai . There are another m cars running in the north-south direction with the i-th car passing the intersection at time bi . If two cars passing the intersections at the same time, a traffic crash occurs. In order to achieve world peace and harmony, all the cars running in the north-south direction wait the same amount of integral time so that no two cars bump. You are asked the minimum waiting time.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 1, 000). The second line contains n distinct integers ai (1 ≤ ai ≤ 1, 000). The third line contains m distinct integers bi (1 ≤ bi ≤ 1, 000).
Output
Print a non-negative integer denoting the minimum waiting time.
Sample Input
1 1
1
1
1 2
2
1 3
Sample Output
1
0

题意：先是东西方向的车子，然后是南北方向的车子
aa数组里面存放的是东西方向的车，bb数组里面存放的是南北方向的车子，东西方向的车子是不会冲突的，然后就看bb数组里面的数是不是与aa里面的数重复，如果重复的话，就要将bb数组全部向后加上一分钟，重新遍历一遍，如果重新遍历的时候，bb的数字还是与aa相冲突就在加一分钟，在重新遍历

#include<iostream>
#include<cstring>
using namespace std;
int aa[12345];int bb[12345];int vis[123456];
int main()
{
	int n,m;
	while(cin>>n>>m)
	{
		int flag=0;
		memset(aa,0,sizeof(aa));
		memset(bb,0,sizeof(bb));
		memset(vis,0,sizeof(vis));
		for(int i=0;i<n;i++) 
		{
			cin>>aa[i];
			vis[aa[i]]=1;
		}
		for(int j=0;j<m;j++) cin>>bb[j];
		for(int i=0;i<m;i++)
		{
			if(vis[flag+bb[i]]) 
			{
				flag++;
				i=0;
			}
		}
		cout<<flag<<endl;
	}
	return 0;
}