# 塔台超频## 题目描述在一条笔直的马路上有 $n$ 个塔台它们被依次标号为 $1 2 cdots n$分别处于距离马路起点 $a _ 1 a _ 2 cdots a _ n$$a _ 1 a _ 2 cdots a _ n$的位置。每个塔台初始时有一个通讯半径 $b _ 1 b _ 2 cdots b _ n$这代表对于 $i$ 号塔台其可以与 $a _ i - b _ i a _ i

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main() {
    int n;
    cin >> n;
    
    vector<pair<int, int>> towers(n);
    for (int i = 0; i < n; i++) {
        cin >> towers[i].first >> towers[i].second;
    }
    
    sort(towers.begin(), towers.end());
    
    int minVoltage = 0;
    int maxRadius = towers[0].second;
    for (int i = 1; i < n; i++) {
        int voltage = max(0, towers[i].first - towers[i-1].first - maxRadius);
        minVoltage = max(minVoltage, voltage);
        maxRadius = max(maxRadius, towers[i].second);
    }
    
    cout << minVoltage << endl;
    
    return 0;
}
``