Codeforces 720C.Cellular Network ( 二分

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本文介绍了一道经典的算法题目——寻找最小覆盖半径，即确保所有城市都能被至少一个信号塔覆盖的最短距离。文章通过二分查找的方法解决该问题，并提供了完整的AC代码实现。

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Cellular Network

题目描述

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.

输入

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.

输出

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

样例

Input
3 2
-2 2 4
-3 0
Output
4
Input
5 3
1 5 10 14 17
4 11 15
Output
3

题意

很裸的二分，注意不要把check函数写挂即可

AC代码

#include <bits/stdc++.h>
using namespace std;

#define LL long long
#define CLR(a,b) memset(a,(b),sizeof(a))

const int MAXN = 1e5+11;
const LL INF = ~0ull>>1;
int arr[MAXN], brr[MAXN];
int n, m;
bool check(int x) {
    int k = 0;
    for(int i = 0; i < m; i++) {
        while(k<n && (abs(brr[i]-arr[k])<=x))
            k++;
    }
    return (k==n);
}
int main() {
    ios::sync_with_stdio(false);
    int ans = -1, res;
    cin >> n >> m;
    for(int i = 0; i < n; i++)
        cin >> arr[i];
    for(int i = 0; i < m; i++)
        cin >> brr[i];
    int l = 0, r = 3e9;
    while(l <= r) {
        int mid = l+(r-l)/2;
        if(check(mid))
            r = mid-1;
        else
            l = mid+1;
    }
    if(!check(r)) r++;
    cout << r << endl;
return 0;
}