codeforces 702C Cellular Network（二分+尺取）

Cellular Network

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.

Examples

Input

3 2
-2 2 4
-3 0

Output

4

Input

5 3
1 5 10 14 17
4 11 15

Output

3

题目大意：有n个城市，m个灯塔，让你求一个最小的灯塔覆盖距离 r ，让n座城市都处在灯塔的照耀下。

解题思路：因为子问题具有独立性，并且答案具有单调性（一个点满足条件，他的左边或是右边都满足条件）。所以采用二分答案的做法来解决这道题。每次二分一个距离 r 。判断这个 r 能否让所有的城市被覆盖。如果可以的话，将其更新为上界，若不行将其更新为下界。中间每一次的判断过程采用尺取法，即当前城市在灯塔的覆盖范围内，继续判断下一个城市，直到出了边界。

二分法注意事项：

1、注意结束条件的写法

2、注意更新边界时区间的开闭

3、判断左右边界谁才是答案

AC代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include<set>
#include<map>
#include<queue>

#define maxn 100005
#define inf 0x3f3f3f3f
#define ONES(x) __builtin_popcount(x)
using namespace std;

typedef long long ll ;
const double eps =1e-8;
const int mod = 1000000007;
typedef pair<int, int> P;
const double PI = acos(-1.0);
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};

int n,m;
int ans;
ll a[maxn],b[maxn];
int main()
{
//freopen("test.txt","r",stdin);
ios_base::sync_with_stdio(false);
cin.tie(0);
int i;
cin>>n>>m;
for(i=0;i<n;i++)
cin>>a[i];
for(i=0;i<m;i++)
cin>>b[i];
ll left = 0,right = 2E9;
ll mid = (left+right)/2;
while(left+1<=right)
{
int num = 0,flag = 0;
for(i=0;i<m;i++)
{
if(b[i]-mid>a[num])
{
flag = 1;
break;
}
while(b[i]+mid>=a[num] && num<n)
num++;
if(num==n)
break;
}
if(num!=n)
flag = 1;
if(flag) left = mid + 1;
else right = mid;
mid = (left+right)/2;
//cout << mid << endl;
}

cout << left << endl;
return 0;
}

/*

unsigned int 0～4294967295
int 2147483648～2147483647
unsigned long 0～4294967295
long 2147483648～2147483647
long long的最大值：9223372036854775807
long long的最小值：-9223372036854775808
unsigned long long的最大值：18446744073709551615

__int64的最大值：9223372036854775807
__int64的最小值：-9223372036854775808
unsigned __int64的最大值：18446744073709551615

*/