codeforces702C Cellular Network(二分)

最新推荐文章于 2020-07-26 22:05:58 发布

T__TSZ

最新推荐文章于 2020-07-26 22:05:58 发布

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本文链接：https://blog.youkuaiyun.com/T__TSZ/article/details/77019135

本文介绍了一种通过二分查找确定最小覆盖半径r的方法，确保所有城市都能被至少一个信号塔覆盖。信号塔为距离其不超过r的城市提供网络服务。文章详细阐述了输入输出格式及示例代码。

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You are given n points on the straight line — the positions (x-coordinates) of the cities and m points on the same line — the positions (x-coordinates) of the cellular towers. All towers work in the same way — they provide cellular network for all cities, which are located at the distance which is no more than r from this tower.

Your task is to find minimal r that each city has been provided by cellular network, i.e. for each city there is at least one cellular tower at the distance which is no more than r.

If r = 0 then a tower provides cellular network only for the point where it is located. One tower can provide cellular network for any number of cities, but all these cities must be at the distance which is no more than r from this tower.

Input

The first line contains two positive integers n and m (1 ≤ n, m ≤ 10⁵) — the number of cities and the number of cellular towers.

The second line contains a sequence of n integers a₁, a₂, ..., a_n ( - 10⁹ ≤ a_i ≤ 10⁹) — the coordinates of cities. It is allowed that there are any number of cities in the same point. All coordinates a_i are given in non-decreasing order.

The third line contains a sequence of m integers b₁, b₂, ..., b_m ( - 10⁹ ≤ b_j ≤ 10⁹) — the coordinates of cellular towers. It is allowed that there are any number of towers in the same point. All coordinates b_j are given in non-decreasing order.

Output

Print minimal r so that each city will be covered by cellular network.

Examples

Input

3 2
-2 2 4
-3 0

Output

Input

5 3
1 5 10 14 17
4 11 15

Output

既然city和Tower是在一条线上求Tower的最小的d，那么可以二分来做

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
long long city[100005];
long long tow[100005];
int main(){
    ios::sync_with_stdio(false);
    int N,M;
    long long high,low,mid,temp,cnt; while(cin>>N>>M)
    {
        for(int i=0;i<N;i++)
            cin>>city[i];
        for(int i=0;i<M;i++)
            cin>>tow[i];
        sort(city,city+N);
        sort(tow,tow+M);
        high=max(city[N-1]-tow[0],tow[M-1]-city[0]);   //上限取city和Tower的最大距离，所以先排了个序
        low=0;
        while(high>low)
        {
            mid=(high+low)/2;
            cnt=0;
            for(int i=0,j=0;j<M&&i<N;i++)
            {
                temp=abs(city[i]-tow[j]);
                if(temp<=mid)
                {
                    cnt++;
                }
                else
                {
                    j++;
                    i--;
                }
            }
            if(cnt<N)
                low=mid+1;
            else
                high=mid;
        }
        cout<<high<<endl;
    }
    return 0;
}