Cf Edu 15 C 城市与信号塔[二分]

题目连接

http://codeforces.com/contest/702/problem/C

Description

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, …, an ( - 109 ≤ ai ≤ 109) — the coordinates of cities. It is allowed that there are any number of cities in the same point. All coordinates ai are given in non-decreasing order.

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

Output

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

Sample Input

5 3
1 5 10 14 17
4 11 15

Sample Output

3

题意

一条直线上有城市和信号塔，保证每个城市都能接受到信号，每个城市与最近的信号塔距离为d，求出最小的ansd，保证所有的城市与最近的信号塔距离都是小于等于ansd的。

题解

对信号塔的位置进行二分，求出与每一个城市最近的信号塔，取其中的最大值，注意特判当信号塔数为1时，取该塔与最左与最右的较大值。

代码

#include<bits/stdc++.h>using namespace std;const int maxn=100005;int a[maxn];int b[maxn];int main(){    int n,m;    scanf("%d%d",&n,&m);    for(int i=0;i<n;i++)    {        scanf("%d",&a[i]);    }    for(int i=0;i<m;i++)    {        scanf("%d",&b[i]);    }    int ans=0;    if(m==1){printf("%d",max(abs(a[0]-b[0]),abs(b[0]-a[n-1])));return 0;}    for(int i=0;i<n;i++)    {        int le,ri,mi,tt;        le=0;ri=m-1;        while(le<ri)        {            mi=(le+ri)/2;            tt=abs(a[i]-b[mi]);            if(b[mi]>a[i])ri=mi;            else le=mi;            if(abs(le-ri)==1)            {                tt=min(abs(a[i]-b[le]),abs(a[i]-b[ri]));                break;            }        }        ans=max(ans,tt);    }    printf("%d\n",ans);    return 0;}